A First Course in Analysis

A First Course In Analysis

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A First Course In Analysis

Donald Yau The Ohio State University at Newark, USA

World Scientific NEW JERSEY

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Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

A FIRST COURSE IN ANALYSIS Copyright © 2013 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

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ISBN 978-981-4417-85-3

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Preface

This is an introductory text on real analysis for undergraduate students. The first course in real analysis is full of challenges, both for the instructors and the students. Many mathematics majors consider real analysis a difficult course. The transition from mechanical computation to formal, rigorous proofs is difficult even for many mathematics majors. Most students beginning a course in real analysis have never been asked to understand and construct proofs before. Moreover, even if one has some ideas about how a proof should go, writing it down in a logical manner is a challenge in itself. This book is written with these challenges in mind. The prerequisite for this book is a solid background in freshman calculus in one variable. The intended audience of this book includes undergraduate mathematics majors and students from other disciplines whose use real analysis. Since this book is aimed at students who do not have much prior experience with proofs, the pace is slower in earlier chapters than in later chapters. In most instances, motivations for new concepts are explained before the actual definitions. For many concepts that have negations (for example, convergence of a sequence), such negations are also stated explicitly. Wherever appropriate we discuss the basic ideas that lead to a proof before the actual proof is given. Such discussion is intended to help students develop an intuition as to how proofs are constructed. There are exercises at the end of each section and of each chapter. Occasionally, some further topics are explored in these additional exercises.

To the Students There are a few things that you should keep in mind as you work through this book. A professor of mine, who I shall not name here, once told me this: Nobody teaches you mathematics. You teach yourself mathematics. You cannot hope to master the materials in this book simply by watching your instructor lecturing. In fact, if you could do that, this book is not for you. I have tried to make the materials as accessible as possible, but you have to do most of the work. You should attempt as many exercises as possible, whether they are assigned homework or not. Expect to vii

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do lots of scratch work as you attempt the exercises. After you have written down a solution to an exercise, read it again and again, and then some more, to see if every step is logical. Expect to think deep and hard as you go through this book. When you read a proof or an example, make sure you understand where each hypothesis is used. Make sure that you understand every single step in a proof. If you get stuck at a certain proof or step, let some time elapse and go back to it later. In particular, do not expect to understand everything the first time you read it. There are many parts in this book that you should read and think through multiple times if you want to master them. Besides this book, the book [Gelbaum and Olmsted (1964)] is highly recommended as a source of many good and exotic examples. Donald Yau

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Contents

Preface 1.

Sets, Functions, and Real Numbers 1.1 1.2 1.3 1.4 1.5 1.6

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Sequences of Real Numbers . . . . Properties of Limits . . . . . . . . . The Bolzano-Weierstrass Theorem Limit Superior and Limit Inferior . Additional Exercises . . . . . . . . .

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Convergence of Series . . Comparison Tests . . . . Alternating Series Test . Absolute Convergence . . Rearrangement of Series Additional Exercises . . .

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Continuous Functions 4.1 4.2 4.3 4.4

1 4 8 17 21 25 29

Series 3.1 3.2 3.3 3.4 3.5 3.6

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Sets . . . . . . . . . . . . . Functions . . . . . . . . . Real Numbers . . . . . . . Mathematical Induction . Countability . . . . . . . . Additional Exercises . . .

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Sequences 2.1 2.2 2.3 2.4 2.5

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62 67 70 72 77 80 83

Limit Points . . . . . . . . . . . . . . . . . . . Limits of Functions . . . . . . . . . . . . . . Continuity . . . . . . . . . . . . . . . . . . . . Extreme and Intermediate Value Theorems ix

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The Derivative . . . . Mean Value Theorem Taylor’s Theorem . . Additional Exercises .

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Integration 6.1 6.2 6.3 6.4 6.5

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Uniform Continuity . . . . . . . . Monotone and Inverse Functions Functions of Bounded Variation . Additional Exercises . . . . . . . .

Differentiation 5.1 5.2 5.3 5.4

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The Integral . . . . . . . . . . . . . . Integration via Tagged Partitions . Basic Properties of Integrals . . . . Fundamental Theorem of Calculus Additional Exercises . . . . . . . . .

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Sequences and Series of Functions 7.1 7.2 7.3 7.4 7.5

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Pointwise and Uniform Convergence Interchange of Limits . . . . . . . . . Series of Functions . . . . . . . . . . . Power Series . . . . . . . . . . . . . . . Additional Exercises . . . . . . . . . .

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157 163 169 175 181

Hints for Selected Exercises

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List of Notations

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Bibliography

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Index

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Chapter 1

Sets, Functions, and Real Numbers

There are two main purposes of this chapter. First, in sections 1.1 and 1.2 we fix some notations and introduce some terminology about sets and functions that will be used in the rest of this book. We also discuss the very useful tool of mathematical induction in section 1.4. Second, basic properties of the real number system are discussed. In section 1.3 the set of real numbers as an ordered field is discussed. All of the ordered field properties of real numbers should already be familiar to the reader. Also, the Completeness Axiom and the Archimedean Property are discussed. In section 1.5 we discuss countable and uncountable sets, both of which are important concepts for real numbers. It is possible, in fact, logically correct, to construct the real number system with all of its well known properties, starting with some set theory axioms. However, we will not take this path in this book, so most of the basic algebraic properties of the real number system are assumed known. The reader who is interested in a rigorous construction of the real number system may consult [Hobson (1907)], [Rudin (1976)], or [Sprecher (1987)].

1.1

Sets

The purpose of this section is to establish some notations and terminology regarding sets. These concepts will be used throughout the rest of this book. More set theory concepts will be introduced in Section 1.5. A good reference for basic set theory is [Halmos (1974)].

1.1.1

Set Notations

By a set S we mean a possibly empty collection of objects, called elements. If x is an element in S and y is not an element in S, we write x ∈ S and y ∈/ S, respectively. The set with zero element is called the empty set and is denoted by ∅. A set with at least one element is said to be non-empty. We sometimes write {x ∶ x ∈ S} to 1

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specific the elements in a set S. Example 1.1. Here are some examples of sets: ● ● ● ● ● ●

1.1.2

The set Z+ = {1, 2, 3, . . .} of positive integers. The set N = {0, 1, 2, . . .} of natural numbers. The set Z = {0, ±1, ±2, . . .} of integers. ∶ m, n ∈ Z, n =/ 0} of rational numbers. The set Q = { m n The set R of real numbers. The set I of irrational numbers. A real number x is said to be irrational if it is not a rational number. Subsets

Suppose that S and T are sets. If every element in S is also an element in T , then we say that S is a subset of T and write S ⊆ T , or sometimes T ⊇ S. So S is not a subset of T if and only if there exists an element x ∈ S that does not belong to T . The phrase if and only if means implications in both directions. So if A and B are two statements, then A if and only if B means both A implies B (i.e., if A is true, then B is true) and B implies A (i.e., if B is true, then A is true). If S is a subset of T and if there exists an element x ∈ T that does not belong to S, then we write S ⊊ T , or sometimes T ⊋ S, and call S a proper subset of T . If S and T have exactly the same elements, then we write S = T and say that they are equal. Example 1.2. We have the following proper subset inclusions: Z+ ⊊ N ⊊ Z ⊊ Q ⊊ R ⊋ I. The only inclusion that is not obvious is the last one. In other words, is there really a real number x that is not a rational number? The √ answer is yes, as we will see in section 1.3. One example of an irrational number is 2 (Theorem 1.4). 1.1.3

Operations on Sets

Let S and T be two sets. There are several ways to build new sets from these two given sets. The union is defined as S ∪ T = {x ∶ x ∈ S or x ∈ T }, an element of which is an element in S or an element in T . For example, we have {0, 2, 5} ∪ {1, 2, 4} = {0, 1, 2, 4, 5}.

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The union of a collection of sets Sn , where n ∈ N, is defined similarly as ∞

⋃ Sn = {x ∶ x ∈ Sn for some n ∈ N}. n=1

The intersection of S and T is defined as S ∩ T = {x ∶ x ∈ S and x ∈ T }. So an element in the intersection S ∩ T is an element that lies in both S and T . For example, we have {0, 2, 5} ∩ {1, 2, 4} = {2}, which consists of the single element 2. The intersection of a collection of sets Sn , where n ∈ N, is defined similarly as ∞

⋂ Sn = {x ∶ x ∈ Sn for every n ∈ N}. n=1

Two sets S and T are said to be disjoint if their intersection S ∩ T is the empty set. If S and T are disjoint, we call S ∪ T their disjoint union. The difference is defined as S ∖ T = {x ∶ x ∈ S and x ∈/ T }. It consists of the elements in S that are not in T . For example, we have {0, 2, 5} ∖ {1, 2, 4} = {0, 5}. The Cartesian product is defined as S × T = {(x, y) ∶ x ∈ S and y ∈ T }. It consists of the ordered pairs whose first entry lies in S and whose second entry lies in T . For example, we have {a, b} × {c, d} = {(a, c), (a, d), (b, c), (b, d)}. The Cartesian product of a collection of sets Sn , where n ∈ N, is defined similarly as ∞

∏ Sn = {(x1 , x2 , . . .) ∶ xn ∈ Sn for each n ∈ N}. n=1

1.1.4

Exercises

In the exercises below, the symbols A, B, C, S, T , etc., denote arbitrary sets. (1) (2) (3) (4)

Are the sets ∅ and {∅} equal? Justify your answer. Prove that S = T if and only if both S ⊆ T and T ⊆ S. Prove that S ⊆ T if and only if S ∩ T = S. Prove the Distributive Laws:

(a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

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(b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). (5) Prove the following generalizations of the Distributive Laws: ∞ (a) A ∪ (⋂∞ i=1 Bi ) = ⋂i=1 (A ∪ Bi ). ∞ (b) A ∩ (⋃i=1 Bi ) = ⋃∞ i=1 (A ∩ Bi ).

(6) Prove the DeMorgan Laws: (a) S ∖ (A ∪ B) = (S ∖ A) ∩ (S ∖ B). (b) S ∖ (A ∩ B) = (S ∖ A) ∪ (S ∖ B). (7) Prove the following generalizations of the DeMorgan Laws: ∞ (a) S ∖ (⋃∞ i=1 Ai ) = ⋂i=1 (S ∖ Ai ). ∞ (b) S ∖ (⋂∞ i=1 Ai ) = ⋃i=1 (S ∖ Ai ). ∞ (8) Prove that S × (⋃∞ n=1 An ) = ⋃n=1 (S × An ). (9) Prove that (A × B) ∩ (S × T ) = (A ∩ S) × (B ∩ T ). (10) For a set S, define its power set P(S) to be the set whose elements are the subsets of S. For example,

P({a, b}) = {∅, {a}, {b}, {a, b}}. If S has n elements, where n is an arbitrary natural number, find the number of elements in its power set P(S).

1.2

Functions

The purpose of this section is to establish some notations and terminology regarding functions. 1.2.1

Functional Notations

The concept of a function should already be familiar from calculus. Let S and T be two sets. A function from S to T , written as f ∶ S → T, is a rule that assigns to each element x ∈ S an element f (x) ∈ T . Equivalently, such a function f is a subset U of the Cartesian product S × T such that for every element x in S, there exists exactly one element in U of the form (x, y) for some y ∈ T . The set S is called the domain of f , which is denoted by Dom(f ). The set T is called the target of f . The range of f is the subset Ran(f ) = {f (x) ∈ T ∶ x ∈ S} ⊆ T. For a subset A of S, the image of A under f is the subset f (A) = {f (a) ∶ a ∈ A} ⊆ T.

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For a subset B of T , the inverse image of B under f is the subset f −1 (B) = {x ∈ S ∶ f (x) ∈ B} ⊆ S. Example 1.3. (1) There is a function f ∶ R → R defined by f (x) = 2x − 1 for real numbers x. The range of f is all of R. The image of Z+ under f is f (Z+ ) = {1, 3, 5, . . .}. The inverse image of Z+ under f is 5 n 3 f −1 (Z+ ) = {1, , 2, , . . .} = {1 + ∶ n ∈ N} . 2 2 2 (2) There is a function g∶ {0, 2, 5} → {1, 2, 4} defined by g(0) = g(2) = 4 and g(5) = 1. The range of g is the proper subset {1, 4} of {1, 2, 4}. Moreover, we have f −1 ({2}) = ∅ and f −1 ({4}) = {0, 2}. 1.2.2

Special Functions

Definition 1.1. Let f ∶ S → T be a function. ● We call f an injection if for elements x and y in S, f (x) = f (y)

implies x = y.

In this case, we also say that f is injective. ● We call f a surjection if for every element z ∈ T , there exists an element x ∈ S such that f (x) = z. In this case, we also say that f is surjective. ● A function that is both an injection and a surjection is called a bijection. Thus, a function f ∶ S → T is not injective if and only if there exist two distinct elements x and y in S such that f (x) = f (y). A function f ∶ S → T is not surjective if and only if there exists an element z ∈ T that does not lie in the range of f . If f ∶ S → T is a bijection, then there is a bijection g∶ T → S (Exercise (13)). In other words, there exists a bijection from S to T if and only if there exists a bijection from T to S. A function can be injective without being surjective, or surjective without being injective. Also, it can be neither injective nor surjective. These statements can be demonstrated with the following examples. Example 1.4. Suppose f ∶ R → R is a function. (1) (2) (3) (4)

If If If If

f (x) = 2x , then it is injective but not surjective. f (x) = x(x − 1)(x + 1), then it is surjective but not injective. f (x) = x2 , then it is neither injective nor surjective. f (x) = 2x − 1, then it is a bijection.

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1.2.3

Inverse Functions

Suppose that f ∶ S → T is an injection. In other words, if x and y are two distinct elements in S, then f (x) =/ f (y) in T . In this case, there is an inverse function f −1 ∶ Ran(f ) → Dom(f ) defined by f −1 (z) = x

if and only if f (x) = z

(1.1)

for all x ∈ S = Dom(f ) and z ∈ Ran(f ). Example 1.5. The function f ∶ R → R defined as f (x) = 2x − 1 is a bijection, which is, in particular, an injection. Its inverse function f −1 ∶ R → R is given by f −1 (x) = 12 (x + 1) for any real number x. If f is a bijection, then its inverse function f −1 is also a bijection. Moreover, the inverse function (f −1 )−1 of f −1 is f . This is Exercise (13) below. 1.2.4

Composition

If f ∶ S → T and g∶ T → U are functions, then their composition g ○ f∶ S → U is defined as (g ○ f )(x) = g(f (x)) for elements x in S. Note that in order to define the composition g ○ f , the range of f must be a subset of the domain of g. Example 1.6. Consider the functions f, g∶ R → R defined by f (x) = x2 and g(x) = 2x − 1. In this case, we can form the compositions (g ○ f )(x) = g(x2 ) = 2x2 − 1 1.2.5

and

(f ○ g)(x) = (g(x))2 = (2x − 1)2 .

Exercises

(1) If f, g∶ S → S are two functions such that f ○ g = g ○ f , does it follow that f = g? (2) Let f ∶ S → T and g∶ T → U be functions. (a) If both f and g are injections, prove that g ○ f is also an injection. (b) If both f and g are surjections, prove that g ○ f is also a surjection. (c) If both f and g are bijections, prove that g ○ f is also a bijection. (3) Let f ∶ S → T and g∶ T → U be functions. (a) If g ○ f is an injection, prove that f is an injection. (b) If g ○ f is an injection, does it follow that g is an injection. (4) Let f ∶ S → T and g∶ T → U be functions.

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(a) If g ○ f is a surjection, prove that g is a surjection. (b) Give an example in which g ○ f is a surjection, but f is not a surjection. (5) For a positive integer n, determine the number of bijections from the set {1, . . . , n} to itself. (6) Let m and n be positive integers. (a) How many different functions f ∶ {1, . . . , n} → {1, . . . , m} are there? (b) Suppose that n ≤ m. How many functions f in the previous part are injections? (7) Let f ∶ S → T be a function, and let A and B be subsets of S. (a) (b) (c) (d)

Prove Prove Prove Is the

that f (A ∪ B) = f (A) ∪ f (B). that f (A ∩ B) ⊆ f (A) ∩ f (B). that, if f is an injection, then f (A ∩ B) = f (A) ∩ f (B). equality f (A ∩ B) = f (A) ∩ f (B) always true?

(8) Let f ∶ S → T be a function, and let A and B be subsets of T . (a) Prove that f −1 (A ∪ B) = f −1 (A) ∪ f −1 (B). (b) Prove that f −1 (A ∩ B) = f −1 (A) ∩ f −1 (B). (c) Prove that f −1 (T ∖ A) = S ∖ f −1 (A). (9) Let f ∶ S → T be a function, and let A1 , A2 , . . . be subsets of T . ∞ −1 (a) Prove that f −1 (⋃∞ n=1 An ) = ⋃n=1 f (An ). ∞ ∞ −1 (b) Prove that f (⋂n=1 An ) = ⋂n=1 f −1 (An ).

(10) Let f ∶ S → T be an injection. Prove that it has a unique inverse function. (11) In each case, (i) verify that f is an injection, (ii) find the inverse function f −1 of f , and (iii) specify domain and range of f −1 . √ (a) f (x) = √x for x ≥ 0. (b) f (x) = 2 + 3x for x ≥ −2/3. (c) f (x) = 1 + x2 for x ≥ 0. (d) f (x) = 2x + 5 for x ∈ R. (12) Let f ∶ S → T and g∶ T → U be injections. Prove that (g ○ f )−1 = f −1 ○ g −1 . (13) Let f ∶ S → T be a bijection. (a) Prove that its inverse function f −1 is also a bijection. (b) Prove that the inverse function (f −1 )−1 of f −1 is f itself. (c) Prove that f −1 (f (x)) = x and f (f −1 (y)) = y for all x ∈ S and y ∈ T . (14) Let f ∶ S → T and g∶ T → U be two functions. Suppose that (g ○ f )(x) = x and (f ○ g)(y) = y for all x ∈ S and y ∈ T . (a) Prove that f and g are both bijections. (b) Prove that g = f −1 .

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1.3

A First Course in Analysis

Real Numbers

In the section, some basic algebraic properties of real numbers are discussed. Then we state the Completeness Axiom and discuss some of its consequences, including the Archimedean Property and characterization of intervals. 1.3.1

Ordered Field

Let x, y, and z be three arbitrary real numbers. There are two basic operations of real numbers. Namely, one can form the sum x + y and the product xy. The following properties should be familiar to the reader. (F1) (F2) (F3) (F4) (F5) (F6) (F7) (F8) (F9)

x + y = y + x. (x + y) + z = x + (y + z). There exists an element 0 such that x + 0 = x. There exists a real number −x such that x + (−x) = 0. xy = yx. (xy)z = x(yz). There exists an element 1 such that 1 ⋅ x = x ⋅ 1. If x =/ 0, then there exists an element x−1 such that x ⋅ x−1 = 1. x(y + z) = xy + xz.

The first two properties state that addition is commutative and associative. The properties (F3)-(F4) state that there exists a 0 real number, and every real number has an additive inverse. The properties (F5)-(F8) are the corresponding statements for multiplication. The last property says that multiplication is distributive over addition. A set S equipped with two operations + (addition) and × (multiplication) satisfying the properties (F1)-(F9) is called a field. So the set R of real numbers is a field, as is the set Q of rational numbers (Exercise (1)). Given two arbitrary real numbers x and y, it is possible to compare them using the inequality ≤ (less than or equal to). The following properties of this relation should be familiar to the reader. Again, x, y, and z are arbitrary elements in R. (O1) Either x ≤ y or y ≤ x; both of these happen at the same time if and only if x = y. (O2) If x ≤ y and y ≤ z, then x ≤ z. (O3) If x ≤ y, then x + z ≤ y + z. (O4) If x ≤ y and 0 ≤ z, then xz ≤ yz. If x ≤ y and x =/ y, then we write x < y and say that x is strictly less than y. We will also use the notations y ≥ x for x ≤ y and y > x for x < y. For real numbers x < 0 < y, we say that x is negative and y is positive.

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A field F equipped with a relation ≤ satisfying the properties (O1)-(O4) is called an ordered field. So the set R of real numbers is an ordered field, as is the set Q of rational numbers. Given a finite number of real numbers {x1 , . . . , xn }, their maximum and minimum elements are denoted by max{x1 , . . . , xn } and min{x1 , . . . , xn }, respectively. 1.3.2

Absolute Value

Another familiar property of real numbers is the absolute value. Definition 1.2. Let x be a real number. Its absolute value is defined as the real number ⎧ ⎪ if x ≥ 0, ⎪x ∣x∣ = ⎨ ⎪ ⎪ ⎩−x if x < 0. Note that ∣x∣ ≤ y

if and only if

− y ≤ x ≤ y.

(1.2)

In particular, applying this to y = ∣x∣, we have −∣x∣ ≤ x ≤ ∣x∣

(1.3)

for any real number x. The most important property of the absolute value that we will use is the Triangle Inequality. Recall from calculus that if a and b are two vectors, one can form a triangle in which the edges are a, b, and a + b. The sum of the lengths of the first two edges, ∣a∣ + ∣b∣, is at least the length of the remaining edge ∣a + b∣. The Triangle Inequality is the real number analog of this fact. Theorem 1.1 (Triangle Inequality). For any real numbers x and y, we have ∣x + y∣ ≤ ∣x∣ + ∣y∣. Proof.

Adding the inequalities −∣x∣ ≤ x ≤ ∣x∣ and

− ∣y∣ ≤ y ≤ ∣y∣,

we obtain −(∣x∣ + ∣y∣) = −∣x∣ − ∣y∣ ≤ x + y ≤ ∣x∣ + ∣y∣. Using the characterization (1.2), the above inequalities are equivalent to the Triangle Inequality. 

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1.3.3

Upper and Lower Bounds

The Completeness Axiom of R is about the existence of a certain upper bound of sets of real numbers. Before discussing this axiom, we first discuss the concepts of upper and lower bounds. Definition 1.3. Let S be a non-empty subset of R. (1) The set S is bounded above if there exists a real number u such that x≤u

for all x ∈ S.

Such an element u is called an upper bound of S. (2) The set S is bounded below if there exists a real number l such that l≤x

for all x ∈ S.

Such an element l is called a lower bound of S. (3) The set S is bounded if it is both bounded above and bounded below. A set is unbounded if it is not bounded. So S is not bounded above if for every real number a, there exists an element x ∈ S with x > a. Likewise, y is not an upper bound of S if and only if there exists an element x ∈ S such that x > y. Similarly, S is not bounded below if for every real number a, there exists an element x ∈ S with x < a. Finally, y is not a lower bound of S if and only if there exists an element x ∈ S such that x < y. Example 1.7. (1) The subset S={

1 ∶ n ∈ Z+ } n

of R is bounded. It is bounded above with 1 serving as an upper bound. Any real number x ≥ 1 is also an upper bound of S. Also, S is bounded below with 0 serving as a lower bound. Any real number y ≤ 0 is also a lower bound of S. In particular, upper and lower bounds are not unique. (2) The subset T = {x ∶ x < 0} of R is bounded above with 0 as an upper bound. However, T is not bounded below. (3) The subset U = {x ∶ x > 1} of R is bounded below with 1 as a lower bound. However, U is not bounded above.

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As noted above, upper and lower bounds, when they exist, are not unique. Thus, when a subset S of R is bounded above, it makes sense to ask if there is an upper bound u that is in some sense the most efficient one. In other words, is there an upper bound u of S such that no real number x < u is an upper bound of S? To make this precise, we need the following definition. Definition 1.4. Let S be a non-empty subset of R. (1) Suppose that S is bounded above. A real number u is called a supremum, or least upper bound, of S if ● u is an upper bound of S, and ● if x < u, then x is not an upper bound of S. We denote a supremum of S by sup(S). (2) Suppose that S is bounded below. A real number l is called an infimum, or greatest lower bound, of S if ● l is a lower bound of S, and ● if x > l, then x is not a lower bound of S. We denote an infimum of S by inf(S). Note that at this point, we do not really know if a set S that is bounded above has a supremum or not. However, if a supremum of S exists, then it must be unique (Exercise (10)). The same is true for infimum, so we can speak of the supremum and the infimum of S. Example 1.8. Consider the subset S = {x ∶ x < 0} of R. It is bounded above with 0 as an upper bound. It is intuitively clear that 0 should be the least upper bound as well. To prove this, pick any real number y < 0. We want to show that y is not an upper bound of S. It suffices to demonstrate the existence of an element x ∈ S with x > y. Since y < 0, we have y < 2y < 0, and so y2 ∈ S shows that y is not an upper bound of S. This shows that sup(S) = 0. Observe that the supremum 0 is not an element in S.

1.3.4

The Completeness Axiom

In this book, we will take the following property of R for granted. The Completeness Axiom. Every non-empty subset S of R that is bounded above has a supremum. So the Completeness Axiom of R guarantees the existence of the supremum as long as the non-empty set is bounded above. Although this axiom is stated only for sets that are bounded above, there is a corresponding statement for sets that are bounded below. It states that every non-empty subset S of R that is bounded below has an infimum (Exercise (9b) below). The Completeness Axiom is extremely important because many results later depend on it.

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We will now discuss some consequences of the Completeness Axiom. Suppose that every step that you take covers a distance of, say, two feet. If you want to get from town A to town B by walking, it seems obvious that you can accomplish this, provided that you walk long enough. This is the motivation for the following result. Theorem 1.2 (The Archimedean Property). Let a and b be positive real numbers. Then there exists a positive integer n such that na > b. Proof. This is proved by contradiction. Suppose that na ≤ b for every positive integer n. So the set S = {na ∶ n = 1, 2, . . .} is not empty, since it contains a, and is bounded above with b as an upper bound. By the Completeness Axiom, the set S has a supremum u. Since a > 0, it follows that u − a < u. With u being the least upper bound of S, this implies that u − a is not an upper bound of S. So there exists an element na ∈ S with na > u − a, or equivalently, (n + 1)a > u. But n + 1 is a positive integer, so (n + 1)a ∈ S and (n + 1)a ≤ u because u is an upper bound of S. This contradicts the statement (n + 1)a > u. Therefore, there exists a positive integer n with na > b.  Given a real number a > 0, it seems clear that the reciprocal of some large positive integer n lies in between 0 and a. The following result shows that this is indeed the case. Corollary 1.1. For every positive real number a, there exists a positive integer n such that n1 < a. Proof. Applying the Archimedean Property to the positive real numbers a and 1, we obtain a positive integer n such that na > 1. The desired inequality now follows when we divide by n.  Another consequence of the Archimedean Property is that the set N of natural numbers is not bounded above. Corollary 1.2. Let x be a real number. Then there exists a positive integer n such that x < n. Proof. If x ≤ 0, then we can take n = 1. So suppose that x > 0. Applying the Archimedean Property to 1 and x, we obtain a positive integer n such that n ⋅ 1 = n > x.  It seems obvious that every positive real number is either a natural number or lies between two natural numbers. This is proved in the following result. Corollary 1.3. Let x be a positive real number. Then there exists a positive integer n such that n − 1 ≤ x < n.

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Proof. set

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By Corollary 1.2 there exists at least one positive integer m > x. So the S = {m ∈ Z+ ∶ m > x}

is non-empty. Let n be the least positive integer in S. Then we have n > x. Also, n − 1 is a natural number that is strictly less than n, so n − 1 ∈/ S and n − 1 ≤ x. Thus, we have n − 1 ≤ x < n.  1.3.5

Lots of Rationals and Irrationals

Next we want to establish the fact that given any two distinct real numbers x and y, there exist a rational number a and an irrational number b, both of which lie strictly between x and y. This is again a consequence of the Archimedean Property. First we prove the rational case. Theorem 1.3. Let x and y be real numbers with x < y. Then there exists a rational number a such that x < a < y. Proof. We prove this when x > 0. The other case x ≤ 0 is Exercise (20). Since y − x > 0, there exists a positive integer n satisfying n1 < y − x by Corollary 1.1. Rearranging this inequality we obtain 1 + nx < ny. Since nx > 0, by Corollary 1.3 there exists a positive integer m such that m − 1 ≤ nx < m. This implies that m ≤ 1 + nx < ny. Thus, we have nx < m < ny. Dividing by n, we obtain m x< < y, n which proves the theorem.

 √ For the irrational case, we will need the following result, which shows that 2 is an irrational number. √ Theorem 1.4. The real number 2 is irrational. √ for some integers Proof. This is proved by contradiction. Suppose that 2 = m n m and n with n =/ 0. Since we can always cancel the common integer factors of m and n, we may √ assume that m and n do not have any common factor > 1. Squaring the equality 2n = m, we obtain 2n2 = m2 . This shows that 2 is a factor of m2 and hence of m. It follows that m2 has at least two factors of 2. The above equality then tells us that n2 must have at least one factor of 2 as well, so n also has a factor of √ 2. Thus, 2 is a common factor of m and n, which is a contradiction. Therefore, 2 is an irrational number. 

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Corollary 1.4. Let x and y be real numbers with x < y. Then there exists an irrational number b such that x < b < y. √ √ Proof. √ Applying √ Theorem 1.3 to√ 2x < 2y, we obtain a rational number a such that 2x < a < 2y. Dividing by 2, we obtain a x < √ < y. 2 Observe that b = √a2 is an irrational number. Indeed, if b is rational, then since a is √  rational, so is 2 = ab , contradicting Theorem 1.4. 1.3.6

Intervals

Many results that we will discuss in later chapters have to do with intervals, which we define formally below. As you already know, if an interval contains two points, then any point in between is also in the interval. Definition 1.5. By an interval in R we mean a non-empty subset I containing at least two real numbers such that if s and t are both in I with s < t, then any real number x satisfying s < x < t is also an element in I. For real numbers a and b with a < b, it is easy to see that the following sets are intervals: (1) (2) (3) (4) (5) (6) (7) (8) (9)

(a, b) = {x ∶ a < x < b} (open bounded interval). (a, b] = {x ∶ a < x ≤ b} (half-open bounded interval). [a, b) = {x ∶ a ≤ x < b} (half-open bounded interval). [a, b] = {x ∶ a ≤ x ≤ b} (closed bounded interval). (a, ∞) = {x ∶ a < x} (open unbounded interval). [a, ∞) = {x ∶ a ≤ x} (closed unbounded interval). (−∞, a) = {x ∶ x < a} (open unbounded interval). (−∞, a] = {x ∶ x ≤ a} (closed unbounded interval). (−∞, ∞) = R (both open and closed interval).

In fact, an interval must be of one of the above forms. We consider the examples of bounded intervals. Proposition 1.1. Let I be an interval that is bounded. Then there exist some real numbers a and b such that I = (a, b), (a, b], [a, b), or [a, b]. Proof. Since I is bounded above and below, it has a supremum b and an infimum a by the Completeness Axiom. Note that a and b may or may not be elements in I. To prove the assertion, it suffices to prove that every real number x satisfying a < x < b is an element in I.

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Pick any positive real number  with  < min{x − a, b − x}. Note that we have a +  < x < b − . Since a +  > a, there exists an element s ∈ I such that s < a +  < x. Likewise, since b −  < b, there exists an element t ∈ I such that t > b −  > x. So we have s < x < t with s and t in I. By the definition of an interval, we conclude that x is an element in I, as desired.  In Exercise (24) below you are asked to show that an unbounded interval must be of one of the remaining five forms. 1.3.7

Exercises

(1) Prove that the set Q of rational numbers, with the usual addition and multiplication, is a field. Are N and Z fields? (2) This exercise is about the field with two elements. Consider the set Z/2 = {0, 1} with two elements. Define its addition and multiplication by 0 + 0 = 0, 0 + 1 = 1 = 1 + 0, 1 + 1 = 0, 0 ⋅ 0 = 0, 0 ⋅ 1 = 0 = 1 ⋅ 0, 1 ⋅ 1 = 1. (a) Prove that Z/2 is a field. (b) Is Z/2 an ordered field? (3) Let a and b be real numbers. (a) Prove that ∣∣a∣ − ∣b∣∣ ≤ ∣a − b∣. (b) Prove that ∣a − b∣ ≤ ∣a∣ + ∣b∣. Suppose that ∣x − a∣ < 2 and that ∣y − a∣ < 2 . Prove that ∣x − y∣ < . Prove that x = 0 if and only if the inequality ∣x∣ <  holds for every  > 0. Suppose that x < y +  for every  > 0. Prove that x ≤ y. Prove that ∣x − a∣ <  if and only if x lies in the open interval (a − , a + ). Prove that a non-empty subset S of R is bounded if and only if there exists a positive real number M such that ∣x∣ < M for every element x in S. (9) Let S be a non-empty subset of R that is bounded below.

(4) (5) (6) (7) (8)

(a) Define the set −S = {−x ∶ x ∈ S}. Prove that −S is bounded above. (b) Prove that sup(−S) = − inf(S), i.e., − sup(−S) is the greatest lower bound of S. Conclude that a non-empty subset of R that is bounded below has an infimum. (10) Let S be a non-empty subset of R. (a) If S is bounded above, prove that S has a unique supremum. (b) If S is bounded below, prove that S has a unique infimum.

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(11) Let S be a set consisting of n elements, where n is some positive integer. (a) Prove that S is bounded. (b) Prove that inf(S) and sup(S) are both elements in S. (12) Let S be a non-empty bounded subset of R. (a) Let u ∈ R be an upper bound of S. Prove that u = sup(S) if and only if for every real number  > 0 there exists an element x ∈ S such that u −  < x. (b) Let l ∈ R be a lower bound of S. Prove that l = inf(S) if and only if for every real number  > 0 there exists an element y ∈ S such that l +  > y. (13) Let S be a non-empty bounded subset of R. Prove that inf(S) ≤ sup(S). (14) Let S ⊆ T be non-empty bounded subsets of R. Prove the inequalities: inf(T ) ≤ inf(S) ≤ sup(S) ≤ sup(T ). So the infimum of a bigger set is smaller, and the supremum of a bigger set is bigger. (15) Let S and T be non-empty bounded subsets of R such that x ≤ y for all x in S and y in T . Prove the following statements. (a) sup(S) ≤ inf(T ). (b) sup(S) = inf(T ) if and only if, for every  > 0, there exist x in S and y in T such that y − x < . (16) Let S and T be non-empty bounded subsets of R. (a) Prove that inf(S ∪ T ) = min{ inf(S), inf(T ) }, the minimum of inf(S) and inf(T ). (b) Prove that sup(S ∪ T ) = max{ sup(S), sup(T ) }, the maximum of sup(S) and sup(T ). (17) Let a and b be real numbers with a < b. (a) Find the infimum and the supremum of the open interval I = (a, b). (b) Do inf(I) and sup(I) lie in I? (c) Repeat the previous two parts with the closed interval [a, b]. (18) Let S be a non-empty subset of R that is bounded above with supremum b. Suppose that b is not an element in S. Prove that for every  > 0, there exist infinitely many elements in S that lie in the interval (b − , b). (19) Find the infimum and supremum of each of the following sets of real numbers. (a) S = {x ∶ 1 < x ≤ 3}. (b) S = {x ∶ x2 − 2x − 3 < 0}. (c) S = {x ∶ x2 − 5 < 0}. (20) Finish the proof of Theorem 1.3 by proving the case x ≤ 0. (21) Let a < b be real numbers. (a) Prove that there exist infinitely many rational numbers in the interval (a, b). (b) Prove that there exist infinitely many irrational numbers in the interval (a, b).

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√ √ √ √ (22) Prove that the following real numbers are all irrational: 3, 3 2, and 2 + 3. (23) Let I and J be two intervals such that the intersection I ∩ J contains at least two distinct real numbers. Prove that I ∩ J is also an interval. (24) Let I be an interval that is an unbounded set. (a) If I is bounded below, prove that I has the form (a, ∞) or [a, ∞). (b) If I is bounded above, prove that I has the form (−∞, a) or (−∞, a]. (c) If I is neither bounded above nor bounded below, prove that I = R. (25) In each case, write down an explicit bijection between the given sets. (a) (1, 3) and (5, 12). (b) R and (0, 1). (c) [0, 1) and [1, ∞). (26) Describe each of the following sets: ⋂∞ n=1 [n, ∞). 1.4

⋂∞ n=1 [1 −

1 , 1], n

⋂∞ n=1 (1 −

1 , 1), n

and

Mathematical Induction

Many proofs in this book and elsewhere use a technique called mathematical induction. This method allows one to establish the validity of many related statements at the same time. 1.4.1

Induction

We begin by stating a property of the set Z+ of positive integers. If S is a non-empty subset of Z+ , then a least element of S is an element x ∈ S such that x ≤ y for all elements y ∈ S. Well-Ordering Property. Any non-empty subset of Z+ has a least element. This seems rather obvious. If S is a non-empty subset of Z+ , just pick its least element! In any case, we take the Well-Ordering Property as an axiom. Principle of Mathematical Induction. For each positive integer n, let P (n) be a statement. Suppose that the following two conditions hold: (1) P (1) is true. (2) For each positive integer k, if P (k) is true, then P (k + 1) is true. Then P (n) is true for all positive integers n. Proof. Let S be the subset of Z+ consisting of those positive integers n for which P (n) is not true. If S is the empty-set, then we are done. Otherwise, S is not empty. By the Well-Ordering Property there is a least element N in S. Since P (1)

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is true, we have that 1 ∈/ S and that N > 1. So N − 1 is a positive integer that does not lie in S, which means that P (N − 1) is true. By the second hypothesis above, it follows that P (N ) = P ((N − 1) + 1) is true. So N ∈/ S, which is a contradiction. Therefore, S must be empty, and P (n) is true for all positive integers n.  In some situations, the statements P (n) do not start with n = 1 but some other positive integer n0 . In this case, the initial case P (1) should be replaced by P (n0 ) and k should be replaced by k ≥ n0 . We illustrate how to use induction with a few examples. Theorem 1.5 (Bernoulli’s Inequality). Suppose that a > −1 and a =/ 0. Then for integers n > 1, we have (1 + a)n > 1 + na. Proof.

This is proved by induction on n. For the first case, when n = 2, we have (1 + a)2 = 1 + 2a + a2 > 1 + 2a.

Now suppose that (1 + a)n > 1 + na for some n ≥ 2. For the case n + 1, we have (1 + a)n+1 = (1 + a)n (1 + a) > (1 + na)(1 + a) = 1 + (n + 1)a + na2 > 1 + (n + 1)a, as desired.



For a positive integer n, define n factorial as the product n! = 1 ⋅ 2 ⋅ 3⋯(n − 1)n. For example, 1! = 1, 3! = 6, and 5! = 120. We set 0! = 1. For non-negative integers n and k with n ≥ k, define the binomial coefficient as n n! . ( )= k!(n − k)! k

(1.4)

For example, n n 4 5 ( ) = 1, ( ) = 1, ( ) = 6, and ( ) = 10. 0 n 2 2 The binomial coefficients are related as follows. The following result is proved without induction. It is needed for the result after it. Its proof is a simple algebraic computation, which is left to the reader as an exercise. Theorem 1.6 (Pascal’s Triangle). For positive integers n and k with n ≥ k, we have n n n+1 ( )+( )=( ). k k−1 k

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Theorem 1.7. For a real number x and a positive integer n, we have n n (1 + x)n = ∑ ( )xk . k=0 k

Proof.

This is proved by induction on n. The first case is 1 1 (1 + x)1 = 1 + x = ( )x0 + ( )x1 . 0 1

Suppose that the required identity is true for some positive integer n. We must show that the next case is true. We compute as follows: (1 + x)n+1 = (1 + x) ⋅ (1 + x)n n n = (1 + x) ⋅ ∑ ( )xk k=0 k n n n n = ∑ ( )xk + ∑ ( )xk+1 k k=0 k k=0 n n n n n )xk + ( )xn+1 = 1 + ∑ ( )xk + ∑ ( k − 1 k n k=1 k=1 n n n )) xk + xn+1 = 1 + ∑ (( ) + ( k k−1 k=1

=(

n+1 0 n n+1 k n + 1 n+1 )x + ( )x + ∑ ( )x k 0 n+1 k=1

n+1 n+1 k )x . = ∑( k k=0

In the next-to-the-last step, we used Theorem 1.6. The theorem is proved.



Corollary 1.5 (Binomial Theorem). For real numbers a and b and a positive integer n, we have n n (a + b)n = ∑ ( )ak bn−k . k=0 k

Proof. If b = 0 then the identity is clearly true, since n − k > 0 except for n = k. So suppose that b =/ 0. In this case, we have a n (a + b)n = bn (1 + ) b n n a k = bn ∑ ( ) ( ) b k=0 k n n = ∑ ( )ak bn−k , k=0 k

where we used Theorem 1.7 in the second equality.



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1.4.2

Strong Induction

Sometimes the following form of induction is more convenient to use. Principle of Strong Induction. For each positive integer n, let P (n) be a statement. Suppose that the following two conditions hold: (1) P (1) is true. (2) For each positive integer k, if P (1), . . . , P (k) are true, then P (k + 1) is true. Then P (n) is true for all positive integers n. Proof. This is proved by contradiction. Suppose that P (n) is not true for some n. By the Well-Ordering Property we can choose the least N such that P (N ) is not true. Since P (1) is true, we have that N > 1 and that P (1), . . . , P (N − 1) are true. By the second hypothesis it follows that P (N ) is true, which is a contradiction. Thus, P (n) is true for every positive integer n.  1.4.3

Exercises

for every positive integer n. (1) Prove that ∑nk=1 k = n(n+1) 2 n(n+1)(2n+1) n 2 (2) Prove that ∑k=1 k = for every positive integer n. 6 2

(3) Prove that ∑nk=1 k 3 = ( n(n+1) ) . 2 n (4) Prove that ∑k=1 (8k − 5) = 4n2 − n for every positive integer n. 2 (5) Prove that ∑nk=1 (2k − 1)2 = n(4n3 −1) for every positive integer n. 1 − xn+1 (6) Prove that ∑nk=0 xk = for every real number x =/ −1 and positive integer 1−x n. n ) for all integers n ≥ k ≥ 0. (7) Prove that (nk) = (n−k n n (8) Prove that 7 − 2 is divisible by 5 for every positive integer n. (9) Prove that 11n − 4n is divisible by 7 for every positive integer n. (10) Prove that 52n − 1 is divisible by 8 for every positive integer n. (11) Prove that ∑nk=1 21k < 1 for every positive integer n. (12) Prove that 2n < n! for all integers n ≥ 4. (13) Prove that n2 < 2n for all integers n ≥ 5. (14) Prove that n3 ≤ 3n for every positive integer n. (15) Let a1 , . . . , an be real numbers with n ≥ 2. (a) Prove that ∣a1 + ⋯ + an ∣ ≤ ∣a1 ∣ + ⋯ + ∣an ∣. (b) Prove that ∣a1 + ⋯ + an ∣ ≥ ∣a1 ∣ − (∣a2 ∣ + ⋯ + ∣an ∣). (16) Let a1 , . . . , an be positive real numbers with n ≥ 1. Prove the inequality: (1 + a1 )⋯(1 + an ) ≥ 1 + a1 + ⋯ + an . (17) Let S1 , . . . , Sn be non-empty bounded subsets of R for some n ≥ 1. Prove that the union ⋃ni=1 Si is also non-empty and bounded. (18) Prove Theorem 1.6.

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1.5

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Countability

Sets come in different sizes. The most basic way to define the size of a set is to categorize it as either finite or infinite. 1.5.1

Finite and infinite sets

Definition 1.6. A set S is called a finite set if either S is the empty set ∅ or there is a bijection from S to the set {1, 2, . . . , n} with n elements for some positive integer n. A set that is not finite is called an infinite set. If there is a bijection f ∶ S → T , then its inverse function f −1 ∶ T → S is also a bijection (Exercise (13) on page 7). Thus, in the above definition, there is a bijection from S to the set {1, 2, . . . , n} if and only if there is a bijection from {1, 2, . . . , n} to S. Example 1.9. The set N of natural numbers is infinite, which seems obvious. To prove it, suppose to the contrary that there exists a bijection f ∶ {1, 2, . . . , n} → N for some positive integer n. We can rank the n natural numbers f (1), . . . , f (n) from the smallest to the largest. If f (k) is the largest integer among these n natural numbers, then f (k) + 1 ∈ N is not in the range of f . This contradicts the assumption that f is a bijection. Therefore, the set N is infinite. It should be intuitively clear that any set that contains an infinite set is itself an infinite set. This is made precise in the following result. Theorem 1.8. Let T be an infinite subset of a set S. Then S is an infinite set. Proof. We prove this by contradiction. So suppose that S is not an infinite set, which means that S is a finite set. By definition there is a bijection f ∶ {1, . . . , n} → S for some positive integer n. We will derive the contradiction that T is a finite set. Let k1 , . . . , ki be the list of all the distinct positive integers in {1, . . . , n} such that f (k1 ), . . . , f (ki ) are elements in T . In particular, we have {f (k1 ), . . . , f (ki )} = T. The function g∶ {1, . . . , i} → T defined by g(j) = f (kj ) ∈ T is injective because f is injective. Moreover, g is surjective because every element in T is of the form f (kj ) for some j = 1, . . . , i. Thus, g is a bijection, which means that T is a finite set. This is a contradiction, and hence S is an infinite set.  Corollary 1.6. Let T be a subset of a finite set S. Then T is a finite set. Proof. This is the contrapositive of Theorem 1.8. Indeed, if T is an infinite set, then S is an infinite set as well, contradicting the assumption. 

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22

A First Course in Analysis

Corollary 1.7. The sets Z, Q, and R are all infinite. Proof. We know from Example 1.9 that N is an infinite set. Since N is a subset of Z, Q, and R, it follows from Theorem 1.8 that they are also infinite sets.  1.5.2

Countable and Uncountable Sets

What is perhaps surprising is that there are very different sizes even among infinite sets. To make this precise, we need the following definitions. Definition 1.7. An infinite set S is said to be countable if there is a bijection from Z+ to S. An infinite set that is not countable is called uncountable. So a set S is uncountable if and only if it is neither finite nor countable. An uncountable set is in some sense a lot bigger than a countable set. At this point, it may seem hard to imagine an infinite set that is uncountable. We will address this issue in a short while. Before that we want to show that the set Q of rational numbers is countable. This might also seem counter-intuitive because Q contains many elements that are not in Z+ . Theorem 1.9. The set Q of rational numbers is countable. in lowest Proof. Every rational number can be written uniquely as a fraction m n terms, in which m and n are integers, n > 0, and that m and n do not have common integer factors > 1. Therefore, we can create a list of the rational numbers as follows: 0 1 −1 2 −2 3 −3 . . . 1 2

− 21

3 2

− 32

5 2

− 52

7 2

...

1 3

− 31

2 3

− 23

4 3

− 43

5 3

...

1 4

− 41

3 4

− 34 ⋱

5 4

− 54

7 4

... ⋮

⋮

In this list the nth row contains the rational numbers m in lowest terms whose n denominator is n. Each rational number appears in this list exactly once. Now we define a function f ∶ Z+ → Q by going down the southwest-to-northeast diagonals. In other words, define First diagonal f (1) = 0, 1 Second diagonal f (2) = , f (3) = 1, 2 1 1 Third diagonal f (4) = , f (5) = − , f (6) = −1, 3 2 and so forth. This function f is surjection because every rational number appears in the above list. Also, f is injective because every rational number appears only

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once in the list. Therefore, f is a bijection from Z+ to Q. This shows that Q is countable.  Next we observe that the set R is uncountable, so intuitively R is a lot bigger as a set than Q. We will concentrate on the open interval (0, 1). Theorem 1.10. The set of real numbers in the open interval (0, 1) is uncountable. Ideas of Proof. The plan is to show that, if (0, 1) is countable, then there is a real number in this interval whose nth digit in its decimal expansion differs from that in the nth term in the countable set (0, 1). This would then be a contradiction. Proof. This is proved by contradiction. Suppose that the set (0, 1) is countable, so there is a bijection f ∶ Z+ → (0, 1). Each real number in (0, 1) has a decimal expansion 0.x1 x2 x3 x4 ⋯ in which each digit xi is between 0 to 9. Using the bijection f , we can list all the elements in (0, 1) as follows: f (1) = 0.a11 a12 a13 a14 ⋯, f (2) = 0.a21 a22 a23 a24 ⋯, f (3) = 0.a31 a32 a33 a34 ⋯, ⋮

⋱

Each digit aij is between 0 and 9. To derive a contradiction, we will write down explicitly a real number b in (0, 1) that is different from f (n) for every n ∈ Z+ . Consider the digits along the diagonal: a11 , a22 , a33 , a44 , etc. For each integer n ≥ 1 define the digit ⎧ ⎪ ⎪2 if ann =/ 2, bn = ⎨ ⎪ ⎪ ⎩5 if ann = 2. Consider the real number b = 0.b1 b2 b3 ⋯ ∈ (0, 1). For each integer n ≥ 1, b is different from f (n) = 0.an1 an2 an3 ⋯ because their nth digits, bn and ann , are different. This shows that f is not surjective, contradicting the assumption that f is a bijection. Therefore, (0, 1) is an uncountable set.  The technique used in the proof above is called Cantor’s diagonal argument.

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A First Course in Analysis

1.5.3

Exercises

(1) Let S1 , . . . , Sk be finite sets for some positive integer k. (a) Prove that the finite union ⋃ki=1 Si is also a finite set. (b) Prove the the finite Cartesian product ∏ki=1 Si is also a finite set. (2) Let Sn , where n ∈ Z+ , be an infinite collection of sets. (a) Is the union ⋃∞ n=1 Sn necessarily an infinite set? (b) Is the Cartesian product ∏∞ n=1 Sn necessarily an infinite set? (3) Let S be an infinite set, and let A be a finite set. Prove that S ∖A is an infinite set. (4) Prove that the set N is countable. (5) Let A be a set, and let B be a countable set. Prove that A is countable if and only if there exists a bijection from A to B. (6) Let S be an infinite set. Prove that S has a countable subset. (7) Prove that the set Z of integers is countable by writing down an explicit bijection from Z+ to Z. (8) Let n ≥ 2 be a positive integer. Prove that Z is the disjoint union of n countable sets. (9) Write down a bijection from the set of even integers to the set Z of integers. (10) Let T be a subset of a countable set S. Prove that T is either finite or countable. (11) Let S and T be countable sets. Prove that there is a bijection from S to S ∪ T . (12) Let f ∶ T → S be an injection in which S is countable. Prove that T is either finite or countable. (13) Let S be a countable set, and let T be a set. Suppose that there is a surjection f ∶ S → T . Prove that T is either finite or countable. (14) Let {Sn } be an infinite collection of countable sets, where n ∈ Z+ . Prove that the union ⋃∞ n=1 Sn is a countable set. (15) Prove that the Cartesian product Z+ × Z+ is a countable set by writing down an explicit bijection from Z+ to Z+ × Z+ . (16) Let S be a countable set, and let T be a finite subset of S. Prove that S ∖ T is countable. (17) Let S be an uncountable set, and let T be a countable set. Prove that S ∖ T is uncountable. (18) Let S be a set that contains an uncountable subset. Prove that S is uncountable. (19) Let S be a set, and let T be an uncountable set. Prove that S∪T is uncountable. (20) Let S be a set, and let T be an uncountable set. Suppose that there is a surjection f ∶ S → T . Prove that S is uncountable. (21) (a) Prove that the set R of real numbers is uncountable. (b) Prove that the set I of irrational numbers is uncountable. (22) Let f ∶ S → T be a bijection in which S is uncountable. Prove that T is

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uncountable. (23) Let a and b be real numbers with a < b. (a) Write down an explicit bijection f ∶ (0, 1) → (a, b). (b) Prove that the interval (a, b) is an uncountable set. (24) Let a and b be real numbers with a < b. (a) Prove that there exist countably many rational numbers in (a, b). (b) Prove that there exist uncountably many irrational numbers in (a, b). 1.6

Additional Exercises

(1) Let f ∶ S → T be an injection with inverse function f −1 ∶ Ran(f ) → Dom(f ) = S. For any subset A of Ran(f ), there seem to be two meanings of the symbol f −1 (A): (i) the inverse image of A under f , or (ii) the image of A under f −1 . Convince yourself that these two interpretations give rise to the same subset of S. ∞ (2) Let Sn = { n − 1, n } for √ positive integers √ n. Prove that ⋃n=1 Sn = N. √ (3) Consider the subset Q( 2) = {a + b 2 ∶ a, b ∈ Q} of R. Prove that Q( 2) is a field in which the addition and multiplication are the ones in R. (4) Let S be a non-empty bounded subset of R, and let a be a real number. Define the sets a + S = {a + x ∶ x ∈ S} and aS = {ax ∶ x ∈ S}. (a) (b) (c) (d)

Prove that a + S and aS are both non-empty and bounded. Prove that inf(a + S) = a + inf(S) and sup(a + S) = a + sup(S). If a > 0, prove that inf(aS) = a inf(S) and sup(aS) = a sup(S). If a < 0, prove that inf(aS) = a sup(S) and sup(aS) = a inf(S).

(5) Let {In } be an infinite collection of intervals, where n = 1, 2, . . .. Suppose that the intersection I = ⋂∞ i=1 Ii contains at least two distinct real numbers. Prove that I is an interval. (6) Let S1 , . . . , Sn be countable sets for some positive integer n. Prove that the Cartesian product S1 × ⋯ × Sn is a countable set. (7) Let S be an uncountable set. Prove that there is a proper subset T of S such that there is a bijection from S to T . (8) Let S be an uncountable set, and let T be a countable set. Prove that there is a bijection from S to S ∪ T . (9) Let S be the set of functions from Z+ to the set {0, 1} with two elements. Prove that S is uncountable. (10) Let S and T be two sets. Their symmetric difference is defined as the set S △ T = {x ∶ x ∈ S or x ∈ T and x ∈/ S ∩ T }. (a) Prove that S △ T = (S ∖ T ) ∪ (T ∖ S). (b) Prove that S = (S △ T ) ∪ (S ∩ T ), where the union is disjoint. (c) Prove that S ∩ (T △ U ) = (S ∩ T ) △ (S ∩ U ).

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A First Course in Analysis

(d) Prove that (S ∪ T ) △ (U ∪ V ) ⊆ (S △ U ) ∪ (T △ V ). Does equality hold in general? (11) In this exercise we construct a bijection f ∶ (0, 1] → (0, 1). For each positive integer n, consider the function defined as fn (x) = 23n − x whose domain is 1 ]. Define f ∶ (0, 1] → (0, 1) by setting f (x) = fn (x) when the interval ( 21n , 2n−1 1 1 x ∈ ( 2n , 2n−1 ]. (a) Sketch the graphs of f1 , f2 , f3 , and f4 . 1 (b) Prove that fn is a bijection from its domain to the interval [ 21n , 2n−1 ). (c) Conclude that f is a bijection. (12) By adapting the ideas of the previous exercise or otherwise, construct a bijection between (0, 1) and [0, 1). (13) Prove that there exist pairwise-disjoint intervals In , where n = 1, 2, 3, . . ., such that the closed interval [0, 1] is the union ⋃∞ n=1 In . Here pairwise-disjoint means Ii ∩ Ij = ∅ whenever i =/ j. (14) A subset S of R is said to be open if for every element x ∈ S, there exists  > 0 such that the open interval (x − , x + ) is a subset of S. Say that S is closed if R ∖ S is open. (a) Prove that ∅ and R are both open and closed. (b) Exhibit a subset S of R that is neither open nor closed. (15) Let a and b be real numbers with a < b. (a) Prove that the intervals (a, b), (a, ∞), and (−∞, a) are open. (b) Prove that the intervals [a, b], [a, ∞), and (−∞, a] are closed. (16) Let S1 , S2 , . . . be subsets of R. (a) Suppose that each Sn is open. Prove that the union ⋃∞ i=1 Si is also open. (b) Suppose that each Sn is closed. Prove that the intersection ⋂∞ i=1 Si is also closed. (17) Let S1 , S2 , . . . be open subsets of R. (a) Prove that the finite intersection S1 ∩ ⋯ ∩ Sn is open. (b) Is it necessarily true that the intersection ⋂∞ i=1 Si is open? (18) Let S1 , S2 , . . . be closed subsets of R. (a) Prove that the finite union S1 ∪ ⋯ ∪ Sn is closed. (b) Is it necessarily true that the union ⋃∞ i=1 Si is closed? (19) Let S be a set. Prove that there is an injection from S to P(S). Recall that the power set P(S) of a set S is the set whose elements are the subsets of S. (20) Let S and T be two sets. (a) Suppose that there is a bijection f ∶ S → T . Prove that there is a bijection from P(S) to P(T ). (b) If P(S) = P(T ), is it true that S = T ?

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(21) Let S be a set. Prove that there is no surjection from S to P(S). (22) Prove that P(Z+ ) is uncountable. (23) For a set S, let Pk (S) be the set of subsets of S with exactly k elements, where k is an arbitrary positive integer. (a) Prove that there is a bijection from S to P1 (S). (b) If S has exactly n elements for some positive integer n, prove that Pk (S), where k ≤ n, has (nk) elements. (24) Let k be a positive integer. (a) Prove that Pk (Z+ ) is countable. (b) Prove that the set of all finite subsets of Z+ is countable. (25) For each positive integer n, let In = [an , bn ] be a bounded closed interval. Suppose that In+1 ⊆ In for each n ≥ 1. (a) Prove that the set A = {an ∶ n ≥ 1} is bounded above. (b) Prove that the set B = {bn ∶ n ≥ 1} is bounded below. (c) Let α = sup(A) and β = inf(B). Prove that α ≤ β and that both α and β lie in ⋂∞ n=1 In . (d) Prove that [α, β] = ⋂∞ n=1 In . If α = β, then [α, β] means the set { α }. In particular, the intersection ⋂∞ n=1 In is non-empty. This is called the Nested Interval Property.

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Chapter 2

Sequences

In this chapter we discuss sequences of real numbers, which the reader should have encountered briefly in calculus. Given a sequence, the main question is whether it converges to a limit or not. A related question concerns the convergence of subsequences. The concepts of limits and convergence will occur many more times later in this book. In section 2.1 the concept of a convergent sequence is introduced. In section 2.2 some basic properties of limits are discussed, including preservation of arithmetic operations and certain types of inequalities. A basic but very important result, the Monotone Convergence Theorem, is proved. Section 2.3 is about the Bolzano-Weierstrass Theorem, which guarantees the existence of a convergent subsequence of a bounded sequence. Using this result, we give another characterization of a convergent sequence as a Cauchy sequence. With the concept of Cauchy sequences, one can establish the convergence of a sequence without first knowing what its limit is. Similar Cauchy criteria for convergence will occur again later in this book. In section 2.4 we discuss limit superior and limit inferior. These concepts provide more information about the possible limits of subsequences of a given sequence. They will also be used in the next chapter when we discuss series.

2.1

Sequences of Real Numbers

2.1.1

Definition of a Sequence

A sequence is a list of real numbers. We will define a sequence formally below. Before that let us consider a few examples. Example 2.1. (1) The sequence 1 1 1 {1, , , , . . .} 2 3 4 has an =

1 n

as its nth term. 29

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A First Course in Analysis

(2) The sequence {−1, 1, −1, 1, . . .} n

has an = (−1) as its nth term. (3) The sequence {a, a2 , a3 , a4 , . . .} has an = an as its nth term, where a is some fixed real number. (4) For a > 0 the sequence 1

1

1

{a, a 2 , a 3 , a 4 , . . .} 1

has as its nth term an = a n . (5) The sequence 1 2 1 3 {(1 + 1)1 , (1 + ) , (1 + ) , . . .} 2 3 has as its nth term an = (1 + n1 )n . (6) The Fibonacci sequence {1, 1, 2, 3, 5, 8, 13, 21, 34, . . .} is defined recursively as a1 = a2 = 1

and

an = an−1 + an−2

for n ≥ 3.

The nth term in the Fibonacci sequence is called the nth Fibonacci number. In other words, for n ≥ 3 the nth Fibonacci number is the sum of the previous two Fibonacci numbers. (7) For a real number r, there is a sequence {sn } whose nth term is the sum sn = 1 + r + r2 + ⋯ + rn . This sequence is usually called a geometric series. In the first example { n1 }, if n is large, then its reciprocal n1 is close to 0. The larger n gets, the closer n1 is to 0. Intuitively this tells us that the sequence { n1 } converges to 0. The only problem is that we do not yet have a rigorous definition of convergence. This will be given in a short while. Definition 2.1. A sequence is a function from the set Z+ of positive integers to R. If f ∶ Z+ → R is such a function, we will usually write f (n) as an and call the list {an } = {a1 , a2 , . . .} a sequence. Since f and {an } determine each other, these two ways of representing a sequence are equivalent. Occasionally, we may have a sequence starting with some ak instead of a1 . For example, if an = (n − 1)−1 then we cannot take n = 1. In such cases, we write {an }∞ n=k for the sequence {ak , ak+1 , ak+2 , . . .}.

or

{an }n≥k

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31

Convergent Sequences

For a sequence {an } to converge to some real number L, what it should mean is this: The sequence {an } can get as close to L as one wants, as long as one is allowed to discard a few terms in {an }. Closeness here means distance from L. So if  > 0 is some small positive real number, being close to L means being an element in the interval (L − , L + ). In other words, given any  > 0, all but a finite number of terms an should lie in (L − , L + ). In formal mathematical language, this is phrased as follows. Definition 2.2. Let {an } be a sequence, and let L be a real number. We say that the sequence {an } converges to L if for every  > 0, there exists a positive integer N such that n≥N

implies

∣an − L∣ < .

(2.1)

In this case, we say that L is the limit of the sequence and that {an } is convergent. If {an } converges to L, we will also write lim an = L,

lim an = L,

n→∞

or

an → L,

and vice versa. A sequence that is not convergent is called divergent. Observe that in Definition 2.2, first  > 0 is given, and then N is chosen to make (2.1) true. The condition (2.1) means that ∣aN − L∣ < ,

∣aN +1 − L∣ < ,

∣aN +2 − L∣ < ,

and so forth. In other words, the an are within  of L for all n sufficiently large. Here is the negation of convergence: A sequence {an } does not converge to L if and only if there exists 0 > 0 such that, for every positive integer N , there exists an integer n≥N

such that ∣an − L∣ ≥ 0 .

Observe that this n is dependent on N . Example 2.2. As discussed above, it should be the case that let  > 0 be given. We want to make the inequality ∣

1 n

→ 0. To prove this,

1 − 0∣ <  n

true for all n sufficiently large. By Corollary 1.2 there exists a positive integer N > 1 . Then for any integer n ≥ N , we have ∣

1 1 1 − 0∣ = ≤ < . n n N

This shows that { n1 } converges to 0.

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Example 2.3. The sequence {(−1)n } = {−1, 1, −1, 1, . . .} should not converge to any real number L, since it keeps alternating between 1 and −1. To prove that it is divergent, let L be any real number, and we show that {(−1)n } does not converge to L. Consider 0 = 21 . For any positive integer N , either ∣1 − L∣ ≥

1 2

or

1 ∣ − 1 − L∣ ≥ . 2

In fact, if neither one of these inequalities is true, then by the Triangle Inequality (Theorem 1.1), 2 = ∣1 − (−1)∣ = ∣(1 − L) + (L − (−1))∣ ≤ ∣1 − L∣ + ∣L − (−1)∣ <

1 1 + = 1, 2 2

which is absurd. Therefore, the sequence {(−1)n } is divergent. 1

Example 2.4. For 0 < a < 1 consider the sequence with an = a n . With the example 1 a = 12 , one can guess that a n → 1. To prove this, consider bn > 0 defined by the equality 1

an =

1 , 1 + bn

1

1

which is possible because 0 < a n < 1. We need an estimate of ∣1 − a n ∣. Observe that 1

∣1 − a n ∣ = ∣1 −

1 bn ∣=∣ ∣ < bn , 1 + bn 1 + bn

since 1+bn > 1. Thus, it suffices to get an estimate for bn . Rearranging its definition, we have 1 = 1n = a(1 + bn )n > a(1 + nbn ) by Bernoulli’s Inequality (Theorem 1.5). Since both a and bn are positive, we can rearrange the above inequality to obtain bn <

1−a . an 1

Suppose that  > 0 is given. We want to show that ∣1 − a n ∣ <  for all n sufficiently large. By Corollary 1.2 there exists a positive integer N such that N>

1−a . a

Then for n ≥ N , we have 1

∣1 − a n ∣ < bn < 1

This shows that a n → 1 when 0 < a < 1.

1−a 1−a ≤ < . an aN

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Example 2.5. For ∣a∣ < 1 consider the sequence {an }. If a = 0, then an = 0 for every n, and we have an → 0. If a =/ 0, observe that ∣a∣ > ∣a∣2 > ∣a∣3 > ⋯. With the example a = 21 , it is not hard to guess that the sequence {an } should converge to 0. To prove this, let  > 0 be given. We can write ∣a∣ =

1 1+b

for some b > 0. We estimate ∣an − 0∣ = ∣a∣n as follows: ∣an − 0∣ =

1 1 1 < < , (1 + b)n 1 + nb nb

where the first inequality uses the Bernoulli’s Inequality (Theorem 1.5). By Corol1 . Then for n ≥ N we have lary 1.2 there exists a positive integer N > b ∣an − 0∣ <

1 1 ≤ < . nb N b

This shows that an → 0 for ∣a∣ < 1. 2.1.3

Divergent Sequences

In Example 2.5, suppose that we consider the sequence {an } with a > 1 instead. In this case, we have 0 < a1 < 1, so a1n → 0. This means that an is large when n is large. In fact, an can be as large as one wants, provided that n is sufficiently large. We make this precise in the following definition. Definition 2.3. Let {an } be a sequence. (1) We say that {an } diverges to ∞ if for every real number M , there exists a positive integer N such that n≥N

implies an > M.

In this case, we write lim an = ∞ or

an → ∞,

and vice versa. (2) We say that {an } diverges to −∞ if for every real number M , there exists a positive integer N such that n≥N

implies an < M.

In this case, we write lim an = −∞ and vice versa.

or

an → −∞,

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A sequence {an } does not diverge to ∞ if and only if there exists a real number M0 such that for every positive integer N , there exists an integer n≥N

such that an ≤ M0 .

We leave it to the reader to formulate what it means for a sequence to not diverge to −∞. The reader should be careful that a divergent sequence does not necessarily diverge to ∞ or −∞, as the following example illustrates. Example 2.6. We saw in Example 2.3 that the sequence {(−1)n } is divergent. However, it does not diverge to ∞ or −∞. In fact, pick M0 = 1. Then an = (−1)n ≤ M0 for all n, so {(−1)n } does not diverge to ∞. A similar argument shows that {(−1)n } does not diverge to −∞. Example 2.7. Recall the Fibonacci sequence {1, 1, 2, 3, 5, 8, 13, 21, 34, . . .} with a1 = a2 = 1

and

an = an−1 + an−2

for n ≥ 3. It appears that {an } diverges to ∞. To prove this, let M be a real number. We want to show that an > M for all n sufficiently large. Looking at the Fibonacci numbers, it should be clear that an ≥ n for n ≥ 5. We prove this by induction, with the first case being a5 = 5. The second case is a6 = 8 > 6. Suppose that ak ≥ k for 5 ≤ k ≤ n for some n ≥ 6. Then we have an+1 = an + an−1 ≥ n + (n − 1) > n + 1. Therefore, by induction we conclude that an ≥ n for all n ≥ 5. Thus, we can choose N to be max{5, M + 1}. Then for n ≥ N we have an ≥ n ≥ N > M, showing that the Fibonacci sequence diverges to ∞. 2.1.4

Uniqueness of Limits

If a sequence {an } is convergent, is it possible that there are two distinct limits? If M and L are two limits of this sequence, then the an are all close to L for all n sufficiently large. But the an are also close to M for all n sufficiently large. This suggests that the two limits L and M must, in fact, be equal. This is proved precisely in the next result. Theorem 2.1. Let {an } be a convergent sequence. Then it has a unique limit. Proof.

Suppose that lim an = L and

lim an = M

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for some real numbers L and M . We want to show that L = M . It is enough to show that ∣L − M ∣ can be made arbitrarily small. By the Triangle Inequality (Theorem 1.1), we have ∣L − M ∣ = ∣(L − an ) + (an − M )∣ ≤ ∣L − an ∣ + ∣an − M ∣.

(2.2)

So for an arbitrary  > 0, to ensure that ∣L − M ∣ < , it is enough to have the righthand side of (2.2) to be less than . Since lim an = L, there exists a positive integer N1 such that  n ≥ N1 implies ∣an − L∣ < . 2 Likewise, since lim an = M , there exists a positive integer N2 such that  n ≥ N2 implies ∣an − M ∣ < . 2 Thus, taking N = max{N1 , N2 }, we have that n ≥ N implies   (2.3) ∣L − M ∣ ≤ ∣L − an ∣ + ∣an − M ∣ < + = . 2 2 Since  > 0 is arbitrary, this shows that L = M .  The argument used in the proof of Theorem 2.1 is called an 2 -argument. This type of argument and its variants will appear many more times in this book. In this type of argument, the quantity to be estimated, ∣L − M ∣ in the case above, is split into two parts as in (2.2). An estimate of 2 is obtained for each of the two parts. Then the results are combined as in (2.3) to get the desired estimate. Variations of this argument (e.g., an 3 -argument) involve splitting the quantity to be estimated into three or more parts. 2.1.5

Exercises

(1) In Definition 2.2 suppose that we replace “n ≥ N ” by “n > N ” in (2.1). Show that the resulting definition of convergence is equivalent to the original one. (2) Prove the following statements using the definition of convergence. 3 (a) 2 + → 2. n √ 5− n → 0. (b) √n 7 √ 3 − → 3. (c) n 13 → 4. (d) 4 + √ 3 n 2n (e) → 0, where n! = 1 ⋅ 2⋯(n − 1) ⋅ n. n! (3) Prove the following statements using the definition of convergence. 1 − 4n → 0. (a) n2

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−2 3 − 2n → . 1 + 5n 5 2 2n + 7 2 (c) → . 3n2 − n − 1 3 1 n2 − 15 → . (d) 4n2 + 17 4 (4) Prove the following statements using Definition 2.3. (b)

(a) n2 − 2n − 10 → ∞. (b) 3n → ∞. n! (c) n → ∞. 3 (d) − ln(n) → −∞. (5) Let a be a real number. Prove that the sequence with an = a for all n is convergent with a. √ limit √ (6) Prove that ( n + 1 − n) → 0. 1 (7) Let a > 1 be a real number. Prove that a n → 1. (8) Prove that an → L if and only if (an − L) → 0. (9) Prove that an → L if and only if −an → −L. (10) Prove that an → L if and only if for every  > 0, the interval (L−, L+) contains all but a finite number of the an . (11) Prove that an → 0 if and only if ∣an ∣ → 0. (12) If an → L, prove that ∣an ∣ → ∣L∣. Is the converse true? (13) Suppose that an → L for some real number L > 0. Prove that there exists a positive integer N such that n ≥ N implies an > 0. (14) Suppose that an ≥ b for each n and that an → L. (a) Prove that L ≥ b. (b) If the hypothesis an ≥ b is replaced by an > b, is it necessarily true that L > b? (15) Suppose that ∣an+1 − L∣ < ∣an − L∣ for each integer n ≥ 1. Does it follow that an → L? ∞ (16) Consider the sequences {an }∞ n=1 and {an }n=k , where k is some fixed positive ∞ integer > 1. Prove that {an }n=1 converges to L if and only if {an }∞ n=k converges to L. (17) Suppose that an → L and bn → L (the same L). (a) Prove that (an − bn ) → 0. (b) Prove that (an + bn ) → 2L. (18) Suppose that ∣an ∣ ≤ M for all n, where M is some positive real number. Let p be a positive integer. Prove that annp → 0. (19) Suppose that an → 0 with each an ≥ 0 and that {bn } is a sequence that satisfies ∣bn − L∣ ≤ an for all n ≥ N for some positive integer N . (a) Prove that bn → L.

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(b) Suppose that the hypothesis ∣bn − L∣ ≤ an is replaced by ∣bn − L∣ ≤ can for some fixed positive real number c. Prove that bn → L. (20) Suppose that {an } is a convergent sequence, and let  > 0 be a real number. (a) Prove that there exists a positive integer N such that ∣an − an+1 ∣ <  for all n ≥ N. (b) Prove that there exists a positive integer N such that n, m ≥ N implies ∣an − am ∣ < . (21) Let {an } be a sequence. (a) Write down what it means for {an } to not diverge to −∞. (b) Give an example of a divergent sequence that does not diverge to −∞. (22) (23) (24) (25)

If an → ∞ or −∞, prove that {an } is divergent. Suppose that each an > 0. Prove that an → 0 if and only if Let a > 1 be a real number. Prove that an → ∞. Suppose that an → ∞ and bn → ∞.

1 an

→ ∞.

(a) If c > 0 is a real number, prove that can → ∞. (b) Prove that (an + bn ) → ∞. (c) Prove that an bn → ∞. (26) In each case, construct sequences {an } and {bn } with the given properties. (a) (b) (c) (d)

an → ∞, an → ∞, an → ∞, an → ∞,

bn → ∞, bn → ∞, bn → ∞, bn → ∞,

and (an − bn ) → ∞. and (an − bn ) → 17. bn =/ 0 for each n, and bn =/ 0 for each n, and

an bn an bn

→ ∞. → 17.

(27) Suppose that an → L and that bn → ∞. Prove that (an + bn ) → ∞. 2.2

Properties of Limits

In this section we discuss several basic but important properties of sequences, including boundedness and the Monotone Convergence Theorem 2.3. 2.2.1

Bounded Sequences

If a sequence {an } converges to a limit L, then all but a finite number of the an are within, say, 1 of L. The other terms, say, a1 , . . . , aN −1 , all lie inside a finite interval. Thus, by enlarging this interval to include (L − 1, L + 1), it seems that the entire sequence lies inside some finite closed interval. To make this precise, first we need some definitions. Definition 2.4. A sequence {an } is said to be bounded if it is bounded as a set (Definition 1.3).

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The discussion before the above definition suggests that a convergent sequence is bounded. This is, indeed, the case, as we now show. Theorem 2.2. A convergent sequence is bounded. Proof. Let {an } be a convergent sequence with limit L. Given  = 1, there exists a positive integer N such that n≥N

implies ∣an − L∣ < 1.

This is equivalent to an ∈ (L − 1, L + 1). Then each am satisfies the inequalities min{a1 , . . . , aN −1 , L − 1} ≤ am ≤ max{a1 , . . . , aN −1 , L + 1}, showing that the sequence {an } is bounded.



The converse of the above theorem is not true. In other words, a bounded sequence is not necessarily convergent. Example 2.8. The sequence {(−1)n } is bounded, since the set {−1, 1} is bounded above by 1 and bounded below by −1. However, this sequence is divergent (Example 2.3). 2.2.2

Monotone Sequences

The above theorem and example suggest a natural question: If boundedness itself is not enough to guarantee convergence, is there some additional hypothesis on a bounded sequence that can guarantee its convergence? The answer is yes, and the relevant concepts are given in the definitions below. Definition 2.5. Let {an } be a sequence. (1) The sequence {an } is said to be increasing if an ≤ an+1

for all

n.

an < an+1

for all

n.

It is strictly increasing if

(2) The sequence {an } is said to be decreasing if an ≥ an+1

for all

n.

for all

n.

It is strictly decreasing if an > an+1

(3) The sequence {an } is said to be monotone if it is either increasing or decreasing. It is strictly monotone if it is either strictly increasing or strictly decreasing.

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A sequence {an } is not increasing if and only if there exists an integer n0 such that an0 > an0 +1 . It is not strictly increasing if there exists an integer n0 such that an0 ≥ an0 +1 . It is an exercise for the reader to formulate the negations of decreasing and strictly decreasing. Example 2.9. (1) (2) (3) (4) (5)

The sequence {(−1)n } is neither increasing nor decreasing. The sequence { n1 } is strictly decreasing. The sequence {1 − n1 } is strictly increasing. The sequence {1, 1, 21 , 21 , 31 , 31 , . . .} is decreasing but not strictly decreasing. The Fibonacci sequence {1, 1, 2, 3, 5, 8, 13, . . .} is increasing but not strictly increasing. The following result is an answer to the question above.

Theorem 2.3 (Monotone Convergence Theorem). Let {an } be a monotone sequence. Then {an } is convergent if and only if it is bounded. In particular, a monotone bounded sequence is convergent. Ideas of Proof. The more precise claim and what the proof below shows is that a bounded increasing sequence {an } converges to sup{an ∶ n ∈ Z+ }. Likewise, a bounded decreasing sequence converges to inf{an ∶ n ∈ Z+ }. Proof. If {an } is convergent, then it is bounded by Theorem 2.2. This proves the “only if” direction. For the other direction, suppose that {an } is bounded. We consider the case when {an } is increasing, leaving the decreasing case to the reader as an exercise. Since the non-empty set A = { a1 , a2 , . . . } is bounded above, it has a supremum α by the Completeness Axiom (section 1.3.4). We show that the sequence converges to α = sup(A). So let  > 0 be given. Since α −  < α, α −  is not an upper bound of the set A. In other words, there exists an aN such that α −  < aN . Then for n ≥ N , we have α −  < aN ≤ an ≤ α < α + , from which we obtain ∣an − α∣ < . This proves that lim an = α. 

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The reader should be careful that the Monotone Convergence Theorem does not assert that a convergent sequence is monotone. Example 2.10. The sequence {(−1)n n1 } = {−1, 21 , − 13 , 41 , . . .} converges to 0. However, it is neither increasing nor decreasing. 1 . We show that it is conExample 2.11. Consider the sequence with an = ∑nk=0 k! vergent using the Monotone Convergence Theorem. First, this sequence is strictly increasing, and hence monotone, because

an+1 = an +

1 > an . (n + 1)!

The sequence is bounded below because each an is positive. It remains to show that it is bounded above. For n ≥ 2 we have 1 1 1 1 + + +⋯+ 1! 2! 3! n! 1 1 1 ≤ 1 + 1 + + 2 + ⋯ + n−1 2 2 2 < 2 + 1.

an = 1 +

This shows that 0 < an < 3 for each n, so the sequence is bounded and monotone. Thus, {an } is convergent by the Monotone Convergence Theorem. Example 2.12. Consider the sequence with an = ∑nk=1 k31k . As in the previous example, the an are all positive, hence bounded below, and the sequence is strictly increasing. To show that it is bounded above, observe that n

n n 1 1 1 ≤ < < 1. ∑ ∑ k k k k=1 k3 k=1 3 k=1 2

an = ∑

Thus, {an } is a bounded increasing sequence. By the Monotone Convergence Theorem, it is convergent. 2.2.3

Arithmetics of Sequences

Next we observe that one can apply the usual arithmetic operations to convergent sequences. Theorem 2.4. Suppose lim an = a and lim bn = b, and suppose c is a real number. Then: (1) (2) (3) (4)

lim(an + bn ) = a + b. lim can = ca. lim an bn = ab. If bn =/ 0 for each n and b =/ 0, then lim abnn = ab .

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Proof. We will prove the last part, which is the most difficult one, and leave the first three parts as exercises. So assume that bn =/ 0 for each n and b =/ 0. First we show that there exists a positive real number K such that ∣bn ∣ > K Given

∣b∣ 2

for all

n.

> 0, since lim bn = b, there exists a positive integer N such that n≥N

implies ∣bn − b∣ <

∣b∣ . 2

This implies that ∣b∣ 2 by Exercise (3a) on page 15. This in turn is equivalent to ∣b∣ − ∣bn ∣ <

∣b∣ < ∣bn ∣ 2

for n ≥ N.

Now K=

∣b∣ 1 min {∣b1 ∣, . . . , ∣bN −1 ∣, } 2 2

satisfies ∣bn ∣ > K for all n. To show that lim abnn = ab , we need to estimate ∣

an a ∣an b − abn ∣ ∣an (b − bn ) + bn (an − a)∣ − ∣= = bn b ∣bn b∣ ∣bn ∣∣b∣ 1 ∣an ∣ ∣b − bn ∣ + ∣an − a∣ ≤ ∣bn ∣∣b∣ ∣b∣

(2.4)

where the last step follows from the Triangle Inequality (Theorem 1.1). Given  > 0, we will use an 2 -argument. Since lim an = a, there exists a positive integer N1 such that ∣b∣ . n ≥ N1 implies ∣an − a∣ < 2 Moreover, since {an } is convergent, it is bounded. So there exists a real number M > 0 such that ∣an ∣ < M for all n. This implies that M ∣an ∣ < ∣bn ∣∣b∣ K∣b∣ for all n. Now, since lim bn = b, there exists a positive integer N2 such that K∣b∣ n ≥ N2 implies ∣b − bn ∣ < . 2M Thus, for n ≥ N = max{N1 , N2 }, it follows from (2.4) that ∣

∣an ∣ K∣b∣ 1 ∣b∣ M K∣b∣  an a − ∣< ⋅ + ⋅ < ⋅ + = . bn b ∣bn ∣∣b∣ 2M ∣b∣ 2 K∣b∣ 2M 2

This shows that lim abnn = ab .



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The above theorem is very useful in computing limits, as illustrated in the following examples. Example 2.13. (1) For any positive integer p, the sequence with an = 1 → 0 (Example 2.2) and an = ( n1 )p . n (2) Consider the sequence with

1 n

→ 0 and

1 n2

converges to 0 because

2n2 + 7 . 3n2 − n − 1

an = Since

1 np

→ 0, we have

an =

2 + n72 2 2+0 2n2 + 7 = = . → 1 1 2 3n − n − 1 3 − n − n2 3−0−0 3

(3) Similarly, the sequence with an =

1 − 4n + 3n2 n2

satisfies 1 − 4n + 3n2 1 4 = 2 − + 3 → 0 − 0 + 3 = 3. n2 n n (4) Let a be a real number with 0 < a < 1. Then the sequence with an =

1

an = −2a n + 5an 1

converges to −2 because a n → 1 (Example 2.4) and an → 0 (Example 2.5). The next result says that, if a sequence {cn } is bounded term-wise between two sequences, both of which converge to the same limit, then so does {cn }. Theorem 2.5 (Squeeze Theorem). Suppose that an ≤ cn ≤ bn

for all

n

and

lim an = L = lim bn

for some real number L. Then lim cn = L. Proof.

This is an 3 -argument. We need to estimate ∣cn − L∣ = ∣(cn − an ) + (an − L)∣ ≤ ∣cn − an ∣ + ∣an − L∣,

where the last inequality uses the Triangle Inequality (Theorem 1.1). So we need to estimate the last two terms above. From the given inequalities we have 0 ≤ cn − an ≤ bn − an . Let  > 0 be given. Since lim an = L there exists a positive integer N1 such that  n ≥ N1 implies ∣an − L∣ < . 3

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Likewise, there exists a positive integer N2 such that  implies ∣bn − L∣ < . 3 Thus, for n ≥ N = max{N1 , N2 }, we have n ≥ N2

∣cn − an ∣ ≤ ∣bn − an ∣ ≤ ∣bn − L∣ + ∣L − an ∣ <

  2 + = . 3 3 3

So for n ≥ N = max{N1 , N2 }, we have ∣cn − L∣ ≤ ∣cn − an ∣ + ∣an − L∣ <

2  + = . 3 3

This shows that lim cn = L.



Example 2.14. The sequence with an = −

cos n n

satisfies

1 1 ≤ an ≤ n n

because −1 ≤ cos x ≤ 1 for every real number x. Since both conclude that an → 0 as well. 2.2.4

1 n

→ 0 and − n1 → 0, we

Exercises

(1) Let {an } be a sequence. (a) Write down what it means for {an } to not be decreasing. (b) Write down what it means for {an } to not be strictly decreasing. (2) Finish the proof of Theorem 2.4 by proving the first three parts. (3) Prove Theorem 2.3 when {an } is a bounded decreasing sequence. (4) Prove that the following sequences converge to 0. (a) an = nn!n . 3 (b) an = (−1)n √ . 3 n+17 √ √ (c) an = n + 17 − n − 5. (5) Let {an } be a monotone sequence. (a) If {an } is not bounded above, prove that an → ∞. (b) If {an } is not bounded below, prove that an → −∞. (6) Let {an } be a bounded sequence. If bn → 0, prove that an bn → 0. (7) Suppose that an → L and bn → M . (a) If an ≤ bn for each n, prove that L ≤ M . (b) Given an example in which an < bn for each n and L = M . (8) Suppose that a ≤ cn ≤ b for each n and that cn → L. Prove that a ≤ L ≤ b. (9) Let {an } be a sequence with an ≥ 0 for each n. Suppose that an → L. √ √ (a) Prove that an → √L. √ 3 (b) Prove that 3 an → L.

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(10) Let a1 = 1 and an+1 =

√

1 + an for n ≥ 1.

(a) Prove that an < 2 for all n. (b) Prove that the sequence {an } is increasing. Conclude that {an } is a convergent sequence. (11) Let {an } and {bn } be convergent sequences. (a) Prove that {∣an + bn ∣} is convergent and that lim ∣an + bn ∣ ≤ lim ∣an ∣ + lim ∣bn ∣. (b) Give an example in which the inequality in the previous part is strict. (c) Prove that {∣an bn ∣} is convergent and that lim ∣an bn ∣ = (lim ∣an ∣) (lim ∣bn ∣) . (12) Let a > 0 be a real number. (a) If 0 < r < 1, prove that arn → 0. (b) If r > 1, prove that arn → ∞. 1

(13) Consider the sequence with an = n n . 1

(a) Let bn = n n − 1. Using Theorem 1.7 prove that 1 n = (1 + bn )n > n(n − 1)b2n . 2 (b) Use the previous part to show that bn → 0. 1 (c) Conclude that n n → 1. 1

(14) Prove that (n + 2) n → 1. (15) Let a be a real number with 0 < ∣a∣ < 1. (a) Define r by ∣a∣ =

1 . 1+r

Prove that n∣a∣n <

2 (n − 1)r2

for every integer n ≥ 2. (b) Prove that n∣a∣n → 0. (16) Let {In } be a sequence of closed and bounded intervals such that In ⊇ In+1 for each n. Use the Monotone Convergence Theorem 2.3 to prove that ⋂∞ n=1 In is non-empty. This is called the Nested Interval Property. 2.3

The Bolzano-Weierstrass Theorem

The first purpose of this section is to discuss subsequences. The main result is the Bolzano-Weierstrass Theorem 2.8, which asserts that every bounded sequence has a convergent subsequence. This theorem is then used to prove a very useful result due to Cauchy, which gives a necessary and sufficient criterion for convergence.

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Subsequences

Consider the sequence {an } = {1, 11 , 2, 21 , 3, 31 , . . .}. It neither converges nor diverges to ±∞. However, if you only pick the first entry and every other entry after that, then the resulting sequence {1, 2, 3, . . .} diverges to ∞. On the other hand, if you only consider the other entries, then the resulting sequence {1, 12 , 31 , . . .} converges to 0. This suggests that something interesting can happen if you only pick certain entries in a sequence. Definition 2.6. Let {an } be a sequence. A subsequence of {an } is a sequence {ank } in which 1 ≤ n1

and nk < nk+1

for all k ≥ 1. In other words, a subsequence {ank } is obtained from {an } by taking only the terms an1 , an2 , etc., in this order. Note that the indices nk have to be strictly increasing, that is, n1 < n2 < n3 < ⋯. Example 2.15. (1) Given any sequence {an }, {an } is itself a subsequence, as are {a1 , a3 , a5 , . . .} and {a3 , a6 , a9 , . . .}. (2) The sequence {(−1)n } has {1, 1, 1, . . .} and {−1, −1, −1, . . .} as two subsequences. (3) For the sequence {1, 11 , 2, 21 , 3, 13 , . . .}, {2, 1, 21 , 31 , 3, . . .} is not a subsequence. If a sequence {an } converges to a limit L, then eventually all the an are close to L. So if {ank } is a subsequence, then the ank are all eventually close to L as well. This is made precise in the following result. Theorem 2.6. Suppose {an } converges to L. Then every subsequence {ank } also converges to L. Proof.

Given  > 0 there exists a positive integer N such that n≥N

implies ∣an − L∣ < .

Since the indexes nk are strictly increasing, we can choose an index nK ≥ N . Then for k ≥ K, we have nk ≥ nK ≥ N This proves that lim ank = L.

and ∣ank − L∣ < . 

Corollary 2.1. If {an } has two convergent subsequences with distinct limits, then {an } is divergent. Proof. Indeed, if {an } is convergent, then by Theorem 2.6 every subsequence converges to the same limit, contradicting the hypothesis. 

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Example 2.16. The sequence 1 1 1 1 1 1 {an } = { , 1 − , , 1 − , , 1 − , . . .} 1 1 2 2 3 3 has a subsequence {1, 21 , 31 , . . .}, which converges to 0. But it also has another subsequence {1 − 1, 1 − 12 , 1 − 31 , . . .}, which converges to 1. Thus the sequence {an } is divergent by Corollary 2.1. 2.3.2

Monotone Subsequences

Most sequences are not monotone. It may come as a surprise that every sequence has a monotone subsequence. To prove this result, we need the following concept. Definition 2.7. Give a sequence {an }, a peak is an entry an such that an ≥ ak

for all

k > n.

In other words, a peak is an entry that is also an upper bound for all the entries after it. Theorem 2.7. Let {an } be a sequence. Then it has a monotone subsequence. Proof. There are two cases. First suppose that the sequence {an } has infinitely many peaks. If an1 , an2 , . . . are the peaks with n1 < n2 < ⋯, then the subsequence {ank } consisting of the peaks is decreasing, and hence monotone. On the other hand, suppose that {an } has only finitely many peaks, say, am1 , . . . , amk . Pick an integer n1 > mk . Since an1 is not a peak, there exists an integer n2 > n1

such that

an1 < an2 .

Since n2 > n1 > mk , an2 is not a peak, so there exists an integer n3 > n2

such that an2 < an3 .

Continuing this way, we obtain an increasing subsequence {ank }.



We obtain the Bolzano-Weierstrass Theorem by combining the above theorem with the Monotone Convergence Theorem. Theorem 2.8 (The Bolzano-Weierstrass Theorem). Every bounded sequence has a convergent subsequence.

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Proof.

Let {an } be a bounded sequence. So there exists a positive real number M such that ∣an ∣ < M for all n. By Theorem 2.7 it has a monotone subsequence {ank }. Since ∣ank ∣ < M for all nk , this subsequence is also bounded. This monotone bounded sequence {ank } is convergent by the Monotone Convergence Theorem 2.3.  The reader should be careful that an unbounded sequence can have a convergent subsequence as well. Example 2.17. The sequence {an } = {1, 11 , 2, 21 , 3, 31 , . . .} is not bounded, but it has a convergent subsequence, namely, { 11 , 21 , 31 , . . .}. 2.3.3

Cauchy Sequences

Recall that the Monotone Convergence Theorem 2.3 tells us that a monotone bounded sequence {an } is convergent. In particular, when using this Theorem to establish convergence, we do not need to know in advance what the limit is. We just need to establish that the sequence is bounded and monotone. But what if you have a sequence that is not monotone? Is there a way to tell whether a sequence is convergent without knowing in advance what its limit might be? There is, indeed, such a criterion, which we now discuss. Definition 2.8. A sequence {an } is called a Cauchy sequence if for every  > 0, there exists a positive integer N such that n, m ≥ N implies ∣an − am ∣ < . A sequence {an } is not a Cauchy sequence if and only if there exists 0 > 0 such that for every positive integer N , there exist integers n, m ≥ N such that ∣an − am ∣ ≥ 0 . Notice the similarity between this definition and Definition 2.2 for a convergent sequence. However, the above definition does not involve a limit. Roughly speaking, a Cauchy sequence is a sequence whose terms are eventually all close to each other. We are going to show that a sequence is convergent if and only if it is a Cauchy sequence. One direction is in the following result. Theorem 2.9. If {an } is a convergent sequence, then it is a Cauchy sequence. Proof. This is an 2 -argument. Let L be the limit of the sequence. Given  > 0 there exists a positive integer N such that  n ≥ N implies ∣an − L∣ < . 2 Thus, for integers n, m ≥ N , we have   ∣an − am ∣ = ∣(an − L) + (L − am )∣ ≤ ∣an − L∣ + ∣L − am ∣ < + = . 2 2 Therefore, {an } is a Cauchy sequence. 

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To show that Cauchy sequences are convergent, we need to work a little harder. The following observation will be needed. Lemma 2.1. Every Cauchy sequence is bounded. Proof. Let {an } be a Cauchy sequence. Given  = 1 there exists a positive integer N such that n, m ≥ N

implies ∣an − am ∣ < 1.

In particular, for integers m ≥ N , we have ∣am ∣ − ∣aN ∣ ≤ ∣aN − am ∣ < 1. So ∣am ∣ < ∣aN ∣ + 1

for all m ≥ N,

which implies ∣an ∣ ≤ M = max{∣a1 ∣, . . . , ∣aN −1 ∣, ∣aN ∣ + 1} for all n. Thus, {an } is a bounded sequence.



Using this lemma we can now show that Cauchy sequences are convergent. Theorem 2.10 (Cauchy Convergence Criterion). Let {an } be a Cauchy sequence. Then it is a convergent sequence. Therefore, a sequence is convergent if and only if it is a Cauchy sequence. Proof. This is again an 2 -argument. By Lemma 2.1 {an } is bounded, so the Bolzano-Weierstrass Theorem 2.8 implies that it has a convergent subsequence {ank }. Let L be the limit of this subsequence. We will show that lim an = L. Given  > 0, since lim ank = L, there exists a positive integer nK such that nk ≥ nK

 implies ∣ank − L∣ < . 2

Since {an } is a Cauchy sequence, there exists a positive integer N ≥ nK such that n, m ≥ N

 implies ∣an − am ∣ < . 2

For integers m ≥ N , we have nm ≥ N ≥ nK . This implies that ∣am − L∣ = ∣(am − anm ) + (anm − L)∣ ≤ ∣am − anm ∣ + ∣anm − L∣ < This shows that the sequence {an } converges to L.

  + = . 2 2 

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We illustrate how the Cauchy Convergence Criterion can be used in the following examples. Example 2.18. Consider the sequence with 1 1 1 an = 1 − + − ⋯ + (−1)n−1 . 2 3 n We want to show that this sequence is convergent by showing that it is a Cauchy sequence, so we need to estimate ∣an − am ∣. If m > n, then 1 1 1 + (−1)n+1 + ⋯ + (−1)m−1 ∣ ∣am − an ∣ = ∣(−1)n n+1 n+2 m 1 1 1 1 1 =∣ −( − − )−( ) − ⋯∣ n+1 n+2 n+3 n+4 n+5 ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ positive

positive

1 . ≤ n+1 Given  > 0 we can choose a positive integer N > 1 − 1. Then for integers n, m ≥ N we have 1 1 1 , }≤ < . ∣am − an ∣ ≤ max { n+1 m+1 N +1 This shows that {an } is a Cauchy sequence. By Theorem 2.10 the sequence {an } is convergent. Example 2.19. Consider the sequence with 1 1 1 an = 1 + 2 + 2 + ⋯ + 2 . 2 3 n As in the previous example, we want to show that this is a Cauchy sequence. For m > n we have 1 1 1 am − an = + +⋯+ 2 (n + 1)2 (n + 2)2 m 1 1 1 < + +⋯+ n(n + 1) (n + 1)(n + 2) (m − 1)m 1 1 1 1 1 1 )+( − ) + ⋯( − ) =( − n n+1 n+1 n+2 m−1 m 1 1 1 < . = − n m n Given  > 0 choose a positive integer N > 1 . Then for n, m ≥ N we have 1 1 1 < . ∣an − am ∣ < max { , } ≤ n m N Thus, {an } is a Cauchy sequence, hence convergent by Theorem 2.10. In Example 2.19, in the third line of the calculation we introduced a sum in which pairs of consecutive terms cancel out, leaving only the first and the last terms. This is sometimes referred to as a telescoping sum, and the method of proof is called a telescoping argument.

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2.3.4

Exercises

(1) Suppose that an → ∞. Prove that every subsequence of {an } also diverges to ∞. Then prove the statement with −∞ instead of ∞. (2) Let {an } and {bn } be two Cauchy sequences. Prove directly from the definition that {an + bn } and {an bn } are Cauchy sequences. (3) Prove directly from the definition that the following sequences are Cauchy sequences. (a) an = (b) an = (c) an =

1 . 2+3n 2n . 5n−7 3n2 −1 . 2n2 +5

(4) Give an example of a divergent sequence {an } such that lim(an+1 − an ) = 0. (5) Let {an } be a Cauchy sequence such that each an is an integer. Prove that there exists a positive integer N such that {am }m≥N is a constant sequence. (6) Let {an } be a divergent sequence, and let L be a real number. Prove that there exist a positive real number 0 and a subsequence {ank } such that ∣ank − L∣ ≥ 0 for all nk . (7) Let {an } be a bounded sequence, and let L be a real number. Suppose that every convergent subsequence of {an } converges to L. Prove that an → L. (8) Let {an } be a Cauchy sequence. Suppose that for every  > 0, there exists an integer n > 1 such that ∣an ∣ < . Prove that an → 0. (9) Consider the sequence with 1 1 1 + +⋯+ . 2! 3! n! Prove directly from the definition that {an } is a Cauchy sequence. (10) Consider the sequence with an = 1 +

1 1 1 + +⋯+ . 2 3 n Prove that {an } is not a Cauchy sequence, so {an } is divergent. (11) Give an example of a divergent sequence {an } such that lim(an+N − an ) = 0 for every positive integer N . (12) Let a1 = 1, a2 = 2, and an = 21 (an−1 + an−2 ) for n ≥ 3. an = 1 +

n−1

(a) For n ≥ 2 prove that an+1 − an = (− 21 ) . (b) Prove that {an } is a Cauchy sequence, hence a convergent sequence. (13) Let a1 and a2 be two distinct real numbers. Define an = 12 (an−1 + an−2 ) for n ≥ 3. Prove that {an } is a Cauchy sequence. (14) Let {an } be a sequence. Suppose that there exists a real number r with 0 < r < 1 such that ∣an+1 − an ∣ ≤ rn for all n. Prove that {an } is a Cauchy sequence. (15) Let {an } be a sequence and C be a real number such that 0 < C < 1. Suppose that ∣an+1 − an ∣ ≤ C∣an − an−1 ∣ for all n ≥ 2. Such a sequence is called a contractive sequence.

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(a) Prove that {an } is a Cauchy sequence, hence a convergent sequence. (b) Is {an } necessarily a Cauchy sequence if C = 1? (16) Let {an } be a bounded sequence with α = inf{an } and β = sup{an }. (a) If α =/ an for any n, prove that there exists a decreasing subsequence of {an } that converges to α. (b) If β =/ an for any n, prove that there exists an increasing subsequence of {an } that converges to β. (17) In each case, construct a sequence {an } with the given properties. (a) (b) (c) (d)

There exist subsequences converging to 1, 2, and 3. For every positive integer M , there exists a subsequence converging to M . There exist subsequences converging to 0 and diverging to ∞ and −∞. There exist subsequences diverging to ∞ and −∞, and for every positive integer M , there exists a subsequence converging to M .

(18) Let {an } be a bounded sequence. For each n define bn = sup{ak ∶ k ≥ n}. (a) Prove that {bn } is a bounded decreasing sequence. In particular, {bn } converges to L = inf{bn } by the Monotone Convergence Theorem 2.3. (b) Prove that there exists a subsequence of {an } that converges to L. 2.4

Limit Superior and Limit Inferior

Theorem 2.6 tells us that, for a convergent sequence, every subsequence has to converge to the same limit. On the other hand, in Example 2.16 we saw that it is possible for a divergent sequence to have convergent subsequences with different limits. A natural question arises. Given a sequence, what can be said about the limits of its convergent subsequences? The purpose of this section is to study this question. 2.4.1

Subsequential Limits

First we need some definitions. Definition 2.9. By an extended real number we mean either a real number or one of the symbols ∞ and −∞. The set of extended real numbers is denoted by R ∪ {±∞}. The usual order of the real numbers naturally extends to the extended real numbers by defining −∞ < a < ∞ for all

a ∈ R.

Example 2.20. For a monotone sequence {an }, the symbol lim an always makes sense as an extended real number. In fact, if {an } is bounded, then it is a monotone

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bounded sequence, which is convergent by the Monotone Convergence Theorem 2.3. If, on the other hand, {an } is not bounded, then either lim an = ∞ or

lim an = −∞

by Exercise (5) on page 43. This discussion still makes sense if the an themselves are extended real numbers. For example, if {an } is increasing and am = ∞ for some m, then ak = ∞ for all k ≥ m and lim an = ∞. Definition 2.10. Let {an } be a sequence. By a subsequential limit of {an } we mean an extended real number L such that there exists a subsequence {ank } with lim ank = L. Example 2.21. (1) If {an } → L, where L ∈ R ∪ {±∞}, then the same is true for every subsequence (Theorem 2.6 and Exercise (1) on page 50). So in this case L is the only subsequential limit of {an }. (2) The sequence with an = (−1)n has a subsequence {1, 1, 1, . . .} with limit 1 and another subsequence {−1, −1, −1, . . .} with limit −1. Using Exercise (5) on page 50, one can see that ±1 are the only subsequential limits of {an }. (3) Consider the sequence {an } = {1, 1, 2, 1, 2, 3, 1, 2, 3, 4, . . .}. For each positive integer m, there exists a subsequence whose entries are all equal to m. So m is a subsequential limit of {an }. Moreover, there is a subsequence {1, 2, 3, 4, 5, . . .} diverging to ∞. By Exercise (5) on page 50 again, these are the only subsequential limits. So the set of subsequential limits of {an } is exactly Z+ ∪ {∞}. Theorem 2.11. Every sequence has at least one subsequential limit. Proof. Given any sequence {an }, it has a monotone subsequence {ank } by Theorem 2.7. Then lim ank exists as an extended real number (Example 2.20).  2.4.2

Size of Subsequential Limits

Theorem 2.11 tells us that it always makes sense to talk about subsequential limits, whether the sequence itself is convergent or not. We now want to discuss how large or how small the subsequential limits of a sequence can get. First we need some definitions. Definition 2.11. Let {an } be a sequence. For each n define the extended real numbers sn = sup{ak ∶ k ≥ n} and in = inf{ak ∶ k ≥ n}.

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The limit superior of {an } is defined as the extended real number lim sup an = lim sn . The limit inferior of {an } is defined as the extended real number lim inf an = lim in . If the set {ak ∶ k ≥ n} is not bounded above, then we take sn = ∞. Likewise, if the set {ak ∶ k ≥ n} is not bounded below, then we take in = −∞. The sequence {sn } of extended real numbers is decreasing because the supremum of a smaller set is smaller or stays the same. Similarly, the sequence {in } is increasing. Therefore, by Example 2.20, lim sup an and lim inf an always make sense as extended real numbers. By a sequence we still mean a sequence of real numbers. We will state it explicitly if we consider monotone sequences of extended real numbers. Example 2.22. (1) For the sequence with an = (−1)n , we have sn = 1 and in = −1 for every n. Thus, we have lim sup an = lim sn = 1

and

lim inf an = lim in = −1.

(2) For the sequence {an } = {1, 1, 2, 1, 2, 3, 1, 2, 3, 4, . . .} considered in Example 2.21 above, we have sn = sup{ak ∶ k ≥ n} = ∞ and in = inf{ak ∶ k ≥ n} = 1. Thus, we have lim sup an = ∞ and

lim inf an = 1.

Notice that, in this case, lim inf an ≤ L ≤ lim sup an for every subsequential limit L of {an }. In each case of the example above, observe the following: (1) There exist a subsequence of {an } converging to lim sup an and a subsequence of {an } converging to lim inf an . In other words, both lim sup an and lim inf an are subsequential limits of {an }. (2) Every subsequential limit L of {an } satisfies lim inf an ≤ L ≤ lim sup an . So lim sup an and lim inf an are, respectively, the largest and the smallest possible subsequential limits of {an }. We now show that these statements about limit superior and limit inferior are, in fact, true for all sequences of real numbers. Theorem 2.12. Let {an } be a sequence, and let L be a subsequential limit of {an }. Then lim inf an ≤ L ≤ lim sup an .

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Proof. Suppose that {ank } is a subsequence of {an } such that lim ank = L. We first want to show that L ≤ lim sup an = s. Since lim sn = s, the same is true for any subsequence of {sn }. In particular, we have lim snk = s. Since ank ≤ snk

for all

k,

the inequality is preserved after taking the limits, i.e., L ≤ s. A similar argument proves the other inequality, lim inf an ≤ L. We leave the details of this case to the reader as an exercise. 2.4.3



Limit Superior and Limit Inferior are Subsequential Limits

Theorem 2.13. Let {an } be a sequence. Then there exist subsequences {ank } and {anl } such that lim ank = lim sup an

and

lim anl = lim inf an .

Ideas of Proof. When lim sup an is a real number, the plan is to choose a subsequence of {an } such that ank is close to a suitable sn . The Squeeze Theorem will then be applied. Proof. We will prove the case about limit superior and leave the other case as an exercise for the reader. First consider the case when lim sup an = lim sn = ∞. We want to construct a subsequence {ank } of {an } that diverges to ∞. Since the sequence of extended real numbers {sn } is decreasing, it follows that sn = ∞ for all n. In particular, the set {ak ∶ k ≥ 1} is not bounded above, so it is possible to choose an an1 > 1. Next choose an an2 > 2 with n2 > n1 , which is possible because the set {ak ∶ k > n1 } is also not bounded above. Continuing this way, we obtain a subsequence with ank > k for each positive integer k. Thus, we have lim ank = ∞ = lim sup an . Next consider the case when lim sup an = −∞. We know that an ≤ sn for each n. Given any real number M , there exists a positive integer N such that n≥N

implies sn < M,

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which in turn implies an ≤ sn < M. So {an }, as a subsequence of itself, diverges to −∞ = lim sup an . The only remaining case is when lim sup an = lim sn = L for some real number L. Since {sn } is decreasing, this implies that there exists a positive integer N such that k≥N

implies sk < ∞.

Since sN − 1 is not an upper bound of {ak ∶ k ≥ N }, there exists an an1 with n1 ≥ N such that sN − 1 < an1 ≤ sN . 1 2

Next, since sn1 +1 − is not an upper bound of {ak ∶ k ≥ n1 + 1}, there exists an integer n2 > n1 such that 1 sn1 +1 − < an2 ≤ sn1 +1 . 2 Continuing this way, we obtain a subsequence {ank } such that 1 snk−1 +1 − < ank ≤ snk−1 +1 (2.5) k for all k ≥ 2. Since sn → L, so does the subsequence {snk +1 } by Theorem 2.6. Taking the limit limk→∞ in (2.5) and using the Squeeze Theorem 2.5, we conclude that ank → L, as desired.  2.4.4

Convergence in Terms of Limit Superior and Limit Inferior

Next we observe that a sequence is convergent, or divergent to ±∞, if and only if its limit superior and limit inferior coincide. Theorem 2.14. Let {an } be a sequence. Then lim an = L ∈ R ∪ {±∞}

if and only if

lim inf an = L = lim sup an .

Proof. First consider the “only if” part. It was observed in Example 2.21 (1) that, when an → L, the same is true for any subsequence, and L is the only subsequential limit of {an }. By Theorem 2.13 both lim inf an and lim sup an are subsequential limits of {an }, so they must both be equal to L. Next assume that lim inf an = L = lim sup an . We want to show that an → L. We will consider the case when −∞ < L < ∞, and leave the other cases (L = ±∞) to the reader as an exercise. There exists a positive integer N such that n≥N

implies

− ∞ < in , sn < ∞,

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where in and sn are the infimum and the supremum of the set {ak ∶ k ≥ n}, respectively. Since −∞ < in ≤ an ≤ sn < ∞ for all n ≥ N , it follows from the Squeeze Theorem 2.5 and lim in = L = lim sn that lim an = L.  2.4.5

Exercises

(1) Find the limit inferior and the limit superior of each of the following sequences. (a) (b) (c) (d) (e)

an = n if n is odd, and an = an = n − n1 . an = (−1)n n. 2 −3 . an = (−1)n 5n2n2 +n−1 √ 2 an = n − 2n − n.

1 n

if n is even.

(2) Finish the proof of Theorem 2.13 by showing that the limit inferior of any sequence {an } is a subsequential limit. (3) Finish the proof of Theorem 2.12 by showing that lim inf an ≤ L for any subsequential limit L of {an }. (4) Finish the proof of Theorem 2.14 by showing that, if lim inf an = lim sup an = L ∈ {±∞}, then an → L. (5) Prove that lim sup(c + an ) = c + lim sup an and lim inf(c + an ) = c + lim inf an for any sequence {an } and real number c. Here we interpret c + ∞ as ∞ and c − ∞ as −∞. (6) Prove that lim inf an ≤ lim sup an for any sequence {an }. (7) Prove the inequalities lim inf an + lim inf bn ≤ lim inf(an + bn ) ≤ lim sup(an + bn ) ≤ lim sup an + lim sup bn . Give an example in which both the first and the third inequalities are strict. (8) Let {ank } be a subsequence of {an }. Prove the inequalities lim inf an ≤ lim inf ank ≤ lim sup ank ≤ lim sup an . (9) Let {an } be a Cauchy sequence. Prove directly (i.e., without using Theorem 2.10) that lim inf an = lim sup an . (10) Suppose that an ≤ bn for every n.

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(a) Prove that lim inf an ≤ lim inf bn . (b) Prove that lim sup an ≤ lim sup bn . (11) Let {an } be a sequence, and let {ln } be a sequence in which each ln ∈ R is a subsequential limit of {an }. Suppose that ln → L ∈ R. Prove that L is a subsequential limit of {an }. (12) Prove that lim sup an = − lim inf(−an ) for any sequence {an }. (13) Prove that a sequence {an } is convergent if and only if it is bounded and has exactly one subsequential limit. (14) Prove that an → 0 if and only if lim sup ∣an ∣ = 0. (15) Prove that a sequence {an } is bounded if and only if lim sup ∣an ∣ < ∞. (16) Let {an } be a bounded sequence. (a) Suppose that lim sup an < M for some real number M . Prove that there exists a positive integer N such that n ≥ N implies an < M . (b) Suppose that lim inf an > m for some real number m. Prove that there exists a positive integer N such that n ≥ N implies an > m. (17) Let {an } be a sequence. (a) Suppose that lim sup an ∈ R. Prove that for every  > 0, there exists a positive integer N such that n ≥ N implies an < (lim sup an ) + . (b) Suppose that lim inf an ∈ R. Prove that for every  > 0, there exists a positive integer N such that n ≥ N implies an > (lim inf an ) − . (18) Let {an } be a bounded sequence, and let r be a positive real number. (a) Prove that lim sup(ran ) = r (lim sup an ). (b) Prove that lim inf(ran ) = r (lim inf an ). (19) Suppose that an → L, a positive real number, and that {bn } is any sequence. Prove that lim sup(an bn ) = L (lim sup bn ). (20) Let {an } and {bn } be bounded sequences of non-negative real numbers. Prove that lim sup(an bn ) ≤ (lim sup an ) (lim sup bn ) . Give an example in which this inequality is strict. (21) Let {an } be a sequence. Prove that lim sup an = inf {sn ∶ n ∈ Z+ }

and

lim inf an = sup {in ∶ n ∈ Z+ } ,

where sn = sup{ak ∶ k ≥ n} and in = inf{ak ∶ k ≥ n}. 2.5

Additional Exercises

(1) Let r be an arbitrary real number. (a) Prove that there exists a strictly increasing sequence of rational numbers {an } such that an → r.

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(b) Prove that there exists a strictly increasing sequence of irrational numbers {an } such that an → r. (c) Repeat the above two parts with strictly decreasing instead of strictly increasing. √ √ (2) Let a1 = 2 and an+1 = 2 + an for n ≥ 1. (a) Prove that an < 2 for all n. (b) Prove that an < an+1 for all n. (c) Conclude that the sequence {an } is convergent. (3) Prove directly from the definition that the following sequences are Cauchy sequences. (a) an = (b) an = (c) an =

2 . 1+n2 3 n ∑k=1 3kk . k n ∑k=1 k22 5k .

(4) Prove that the sequence with 1 1 1 an = 1 + √ + √ + ⋯ + √ n 2 3 is divergent. (5) Let p be a real number with 0 < p < 1. Prove that the sequence with 1 1 1 an = 1 + p + p + ⋯ + p 2 3 n is divergent. (6) If lim a2n−1 = L = lim a2n , prove that lim an = L. (7) Suppose that an → L =/ 0 and that {an bn } is a convergent sequence. Prove that {bn } is convergent. (8) Suppose that an → L and that f ∶ Z+ → Z+ is a bijection. Consider the sequence with bn = af (n) . Prove that bn → L. (9) Let p(x) = cm xm +cm−1 xm−1 +⋯+c0 be a polynomial, so the ci are real numbers, and suppose that an → L. Prove that p(an ) → p(L). ) = L. (10) Suppose that each an > 0 such that lim ( aan+1 n (a) If L < 1, prove that there exist a positive real number r with L < r < 1 and a positive integer N such that aN +k < rk aN for all positive integers k. (b) If L < 1, use the previous part to prove that an → 0. (c) If L > 1, prove that an → ∞. (d) Given an example in which L = 1 and {an } is convergent. (e) Given an example in which L = 1 and {an } is divergent. (11) Let p(x) = cm xm + cm−1 xm−1 + ⋯ + c0 and polynomials with dk =/ 0. Prove that cm ⎧ ⎪ ⎪ ⎪ ⎪ p(n) ⎪ ⎪ dk lim = ⎨0 q(n) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩±∞

q(x) = dk xk + dk−1 xk−1 + ⋯ + d0 be if m = k, if m < k, if m > k.

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(12) Let {an } be a sequence. Define another sequence by a1 + ⋯ + an . bn = n (a) If an → L, prove that bn → L. (b) Given an example in which {bn } is convergent but {an } is divergent. n

(13) Consider the sequence with an = (1 + n1 ) . (a) Using Theorem 1.7 prove that 1 1 1 1 2 an = 1 + 1 + (1 − ) + (1 − ) (1 − ) + ⋯ 2! n 3! n n 1 2 n−1 1 ). + (1 − ) (1 − ) ⋯ (1 − n! n n n (b) Use the previous part to show that the sequence {an } is increasing. (c) Prove that 1 1 an ≤ 1 + 1 + + ⋯ + <3 2! n! for each n ≥ 2. (d) Conclude that {an } is a convergent sequence. (14) Find the limit inferior and limit superior of the sequence whose nth term is 3 + (−1)n (2 − n5 ). Does this sequence converge? (15) Consider two sequences {an } and {bn } that are equal as sets. Does it follow that these two sequences have the same set of subsequential limits? (16) Construct a sequence whose set of subsequential limits is {0} ∪ { n1 ∶ n ∈ Z+ }. (17) Prove that there exists a sequence {an } whose set of subsequential limits is exactly the closed interval [0, 1]. (18) Let a < b be real numbers. (a) Prove that there exists a sequence {rn } in (a, b) such that every rational number in (a, b) occurs in this sequence exactly once. (b) Find the set of subsequence limits of the sequence {rn }. (19) Let {an } be a bounded sequence. (a) Prove that lim sup = L if and only if for every  > 0, the inequality ∣an −L∣ <  holds for an infinite number of an , and only finitely many an > L + . (b) Prove that lim inf = L if and only if for every  > 0, the inequality ∣an −L∣ <  holds for an infinite number of an , and only finitely many an < L − . (20) Prove that the following statements are equivalent. (a) (b) (c) (d)

The Bolzano-Weierstrass Theorem 2.8. The Nested Interval Property (Exercise (16) on page 44). Every Cauchy sequence is convergent. The Completeness Axiom.

In other words, prove that each statement implies the others. It suffices, for example, to prove the implications (a) ⇒ (b) ⇒ (c) ⇒ (d) ⇒ (a).

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Chapter 3

Series

In this chapter we discuss a special kind of sequences, called series. In fact, we already discussed several examples of series in the previous chapter. Examples 2.11, 2.12, 2.18, and 2.19 are all series. In general, a series {sn } has the form sn = a1 + ⋯ + an . As with sequences, the main question about a series is whether it converges. In section 3.1 we discuss two convergence criteria for series due to Cauchy. The first one is the Cauchy Convergence Criterion (Theorem 2.10) for sequences, stated in the context of series. The second one is called the Cauchy Condensation Test. The two most basic types of series, the geometric series and the p-series, are also discussed. Several more tests for convergence of series are discussed in sections 3.2 and 3.3. They include the two Comparison Tests and the Alternating Series Test. The reader should already be familiar with these tests from calculus. In section 3.4 we discuss the related concept of absolute convergence. The Ratio Test and the Root Test are discussed. In section 3.5 we consider what can happen when the terms of a series are rearranged. It turns out that for some convergent series, such a rearrangement may converge to a different limit. To motivate the discussion of series, consider the decimal form of a real number r = n.n1 n2 n3 n4 . . . = n + 0.n1 n2 n3 n4 . . . , where n is an integer and each nk ∈ {0, 1, . . . , 9}. The kth digit nk after the decimal nk point represents the real number 10 k . So writing a real number r in its decimal form is actually representing it as a “sum” n3 n1 n2 + + + ⋯. (3.1) r =n+ 10 102 103 However, this “sum” may go on indefinitely. For example, π = 3.14159... has an unending decimal expansion. In other words, we are really considering the sequence {rk } in which n1 n2 nk rk = n + + 2 +⋯+ k, 10 10 10 and the limit of this sequence is r. As with any other sequence, here is the main question. Are we sure that this sequence converges? Can there be decimal expansions that do not converge? If so, which ones? 61

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A First Course in Analysis

The example and questions above illustrate the need for a rigorous definition and careful analysis of series. With the tools that we will develop in this chapter, it is not hard to show that the sequence {rk } above converges. Of course, series are widely used in the mathematical sciences and are not restricted to decimal expansions.

3.1

Convergence of Series

In this section, we discuss some basic tests for convergence of series. We also discuss some examples of series that are useful for comparing with other series. 3.1.1

Definition of a Series

Definition 3.1. A series is a sequence {sn } in which sn has the form sn = a1 + a2 + ⋯ + an for some sequence {an }. We also write such a series as ∞

∑ an

or

∑ an ,

n=1

and call sn the nth partial sum of the series. A series ∑ an is said to be convergent if the sequence {sn } of partial sums is a convergent sequence. If L is the limit of {sn }, we write ∑ an = L. A series that is not convergent is called divergent. If sn → ±∞, then we say that the series diverges to ±∞ and write ∑ an = ±∞. As in the case of sequences, sometimes the an involved may not start with a1 but some other term ak . For example, we may have an = (log n)−1 , which does not make sense if n = 1. In such cases, we start with sk = ak , and we have sn = ak + ak+1 + ⋯ + an for n ≥ k. We write ∞

∑ an n=k

if we wish to emphasize that the first term in the series is ak . We will sometimes write ∑ an = a1 + a2 + a3 + ⋯ just to exhibit the first few ak in the series.

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63

Cauchy Convergence Criterion for Series

Since a series ∑ an is actually a sequence {sn } of partial sums, any convergence criterion for sequences automatically applies to series. In particular, the Cauchy Convergence Criterion for sequences can be used for series as well. Theorem 3.1. A series ∑ an is convergent if and only if for every  > 0, there exists a positive integer N such that n

n>m≥N

implies

∣ ∑ ak ∣ < . k=m+1

Proof. By Theorem 2.10 the sequence {sn } of partial sums is convergent if and only if for every  > 0, there exists a positive integer N such that n, m ≥ N

implies ∣sn − sm ∣ < .

If n = m then this inequality is trivially true. If n =/ m, then either n > m or m > n. In the case that n > m, we have n

m

n

∣sn − sm ∣ = ∣ ∑ ak − ∑ ak ∣ = ∣am+1 + am+2 + ⋯ + an ∣ = ∣ ∑ ak ∣ . k=1

k=m+1

k=1

In the case that m > n, we simply reverse the roles of n and m in the above discussion.  Taking the negation of the Cauchy Convergence Criterion for series, we obtain a criterion for a series to diverge. Corollary 3.1. A series ∑ an is divergent if and only if there exists 0 > 0 such that, for every positive integer N , there exist integers n

n>m≥N

such that

∣ ∑ ak ∣ ≥ 0 . k=m+1

Another useful criterion for showing that a series diverges is the following result. Corollary 3.2. If the series ∑ an is convergent, then lim an = 0. Equivalently, if {an } does not converge to 0, then the series ∑ an is divergent. Proof.

Given  > 0, by Theorem 3.1 there exists a positive integer N such that n

n>m≥N

implies

∣ ∑ ak ∣ < . k=m+1

In particular, with n = m + 1 > N , we have m+1

∣an − 0∣ = ∣ ∑ ak ∣ < . k=m+1

This shows that an → 0.



Let us emphasize that Corollary 3.2 above is a one-way implication; its converse is false. In other words, lim an = 0 alone is not enough to guarantee the convergence of the series ∑ an . This is illustrated in the following example.

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3.1.3

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A First Course in Analysis

Harmonic and Geometric Series

Example 3.1. The harmonic series is the series 1 1 1 = 1 + + + ⋯. 2 3 n=1 n ∞

∑

Since an = n1 , we have an → 0 (Example 2.2). We now show that the harmonic series is divergent using Corollary 3.1 with 0 = 21 . Given any positive integer N , we can take m = N and n = 2m = 2N . We have 2N

1 1 1 1 1 1 ≥N⋅ = + +⋯+ = . N +1 N +2 2N 2N 2 k=N +1 k ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ∑

N terms

The above inequality holds because each of the N terms is greater than or equal to 1 . This shows that the harmonic series ∑ n1 is divergent, even though n1 → 0. 2N Example 3.2. Let r be a real number. The geometric series is a series of the form ∞ n 2 3 ∑ r = 1 + r + r + r + ⋯. n=0

If ∣r∣ ≥ 1, then ∣r∣n ≥ 1 for all n, and the sequence {rn } does not converge to 0. So by Corollary 3.2 the geometric series ∑ rn diverges when ∣r∣ ≥ 1. When ∣r∣ < 1, the nth partial sum is sn = 1 + r + ⋯ + rn−1 + rn . So we have rsn = r + r2 + ⋯ + rn + rn+1 = sn + rn+1 − 1. Solving for sn in this equality, we have 1 1 1 − rn+1 = − ⋅ rn+1 . 1−r 1−r 1−r From Example 2.5 we know that rn+1 → 0 if ∣r∣ < 1. Thus, we conclude that the 1 n geometric series ∑∞ n=0 r converges to 1−r if ∣r∣ < 1. sn =

The two examples above, the harmonic series and the geometric series, are actually closely related. We will make this clear after the following convergence test. 3.1.4

Cauchy Condensation Test

Theorem 3.2. Suppose that each an ≥ 0 and that the sequence {an } is decreasing. Then the series ∑∞ n=1 an is convergent if and only if the series ∞ k ∑ 2 a2k = a1 + 2a2 + 4a4 + ⋯ k=0

is convergent.

(3.2)

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Ideas of Proof. Both implications are proved using the Monotone Convergence Theorem. Therefore, in each case, we will establish that the sequence of partial sums is bounded and monotone. Proof.

Consider the two sequences of partial sums involved: sn = a1 + a2 + a3 + a4 + ⋯ + an , tn = a1 + 2a2 + 4a4 + 8a8 ⋯ + 2n a2n = a1 + a2 + a2 + a4 + ⋯ + a2n .

It follows from the hypotheses on the ak that 0 ≤ sn ≤ tn

(3.3)

for each n ≥ 1. To prove the “if” part, suppose that the series (3.2) is convergent. In other words, we assume that the sequence {tn } of its partial sums is convergent. So the sequence {tn } is bounded (Theorem 2.2), which implies by (3.3) that the sequence {sn } is bounded. Since ak ≥ 0 for each k, the sequence {sn } is increasing and bounded. The Monotone Convergence Theorem 2.3 now implies that {sn }, and hence the series ∑ an , is convergent. To prove the “only if” part, assume that ∑ an , and hence {sn }, is convergent. Since {sn } is a bounded sequence, there exists M > 0 such that 0 < sn < M for all n. Since the sequence {tn } of partial sums is increasing, to show that it is convergent, it is enough to show that it is bounded. If n = 2k , since the sequence {an } is decreasing, we have 2sn = a1 + a1 + a2 + a2 + a3 + a3 + a4 + a4 +⋯ + an + an ≥ a1 + tk ≥ tk . ´¹¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ ≥2a2

≥4a4

We conclude that tk ≤ 2s2k < 2M for all k. This shows that {tn } is bounded and monotone, and hence the series (3.2) is convergent.  The Cauchy Condensation Test is a remarkable result. It says that the convergence of the series ∑ an is determined by that of a series formed only with the terms a2k (k ∈ N). Example 3.3. Generalizing the harmonic series ∑ n1 (Example 3.1), we consider the p-series 1 1 1 = 1 + p + p + ⋯, p 2 3 n=1 n ∞

∑

where p is a fixed real number. If p ≤ 0, then n−p ≥ 1 for all n, and {n−p } does not converge to 0. So by Corollary 3.2 the p-series ∑ n−p is divergent when p ≤ 0.

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A First Course in Analysis

Suppose that p > 0. In this case, the sequence {n−p } is decreasing and each n−p > 0. Thus, by the Cauchy Condensation Test, the p-series is convergent if and only if the series ∞ n ∑2 ⋅ n=0

1 (2n )p

∞

n

= ∑ [2(1−p) ] n=0

is convergent. This is a geometric series ∑ rn with r = 2(1−p) . By Example 3.2 this geometric series is convergent if and only if 2(1−p) < 1, i.e., p > 1. Therefore, we conclude that the p-series ∑ n−p is convergent if and only if p > 1. 3.1.5

Exercises

(1) Let ∑ an = A and ∑ bn = B be two convergent series, and let c be a real number. (a) Prove that ∑(an + bn ) = A + B. (b) Prove that ∑ can = cA. n (2) Let a and r be real numbers. Prove that the series ∑∞ n=0 ar , also called a a geometric series, is convergent with limit 1−r when ∣r∣ < 1 and is divergent when ∣r∣ ≥ 1. (3) Prove that the harmonic series ∑ n1 diverges to ∞. (4) Determine whether the following series converge.

(a) (b) (c) (d)

∑[log n]−1 . ∑[n log n]−1 . ∑[n(log n)(log log n)]−1 . ∑[n(log n)p ]−1 , where p is a fixed real number.

(5) Prove that the convergence of a series ∑ an is not affected by inserting or deleting finitely many terms. In particular, if ∑∞ n=1 an is convergent, then so a for every positive integer k. is ∑∞ n=k n (6) Let ∑ an be a convergent series. For each m, let bm = ∑∞ n=m an . Prove that bm → 0. (7) Suppose that an → L ∈ R. Prove that the series ∑∞ n=1 (an − an+1 ) converges to a1 − L. −1 (8) Prove that ∑∞ converges to 1. n=1 [n(n + 1)] (9) Suppose that an → ∞. Prove that ∑(an+1 − an ) = ∞. (10) Prove that each of the following series diverges to ∞. √ √ (a) ∑( n + 1 − n). (b) ∑(log(n + 1) − log n). (11) Suppose that the series ∑ ∣an ∣ is convergent. (a) Prove that ∑ an is convergent. (b) Give an example in which ∑ an is convergent but ∑ ∣an ∣ is divergent. (12) Suppose that each an ≥ 0. Prove that the series ∑ an is convergent if and only if its sequence {sn } of partial sums is bounded.

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(13) Prove that the series (3.1) for decimal expansion always converges. (14) Suppose that each an ≥ 0 and that ∑ an is convergent. Suppose that {ank } is a subsequence of {an }. (a) Prove that ∑∞ k=1 ank is convergent. (b) Is ∑ ank still convergent if we do not assume that an ≥ 0 for all n? (15) Suppose that ∑ an is a convergent series with limit S. Define b1 = a1 + ⋯ + an1 , b2 = an1 +1 + ⋯ + an2 , b3 = an2 +1 + ⋯ + an3 , etc., where 1 ≤ n1 < n2 < n3 < ⋯. Prove that ∑ bn converges to S. This says that in a convergent series, if parentheses are introduced to form a new series, then the resulting series converges to the same limit. (16) Prove that the converse of the previous exercise is false. In other words, exhibit an example in which ∑ bn is convergent but ∑ an is divergent. (17) Suppose that each an ≥ 0 and that ∑ an is a convergent series with limit S. Let f ∶ Z+ → Z+ be a bijection, and define bn = af (n) . Prove that ∑ bn = S. This says that if the terms of a convergent series with non-negative terms are rearranged in any order, then the resulting series converges to the same limit. 3.2

Comparison Tests

In this section we discuss two Comparison Tests for convergence of series. 3.2.1

Comparison Test

To motivate the first comparison test, consider the series 1 1 1 1 = + + + ⋯. 2+5 2n 7 13 23 n=1 ∞

∑

Since n2 < 2n2 + 5, taking the reciprocals we have 1 1 0< 2 < 2. 2n + 5 n −2 Since the bigger series ∑ n is a convergent p-series (Example 3.3), it seems that the smaller series ∑(2n2 + 5)−1 should be convergent as well. This is, indeed, the case, as the following test shows. Theorem 3.3 (Comparison Test). Suppose that 0 ≤ an ≤ bn

for all

n.

(1) If ∑ bn is convergent, then ∑ an is convergent. (2) If ∑ an is divergent, then ∑ bn is divergent.

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Proof.

A First Course in Analysis

Consider the sequences of partial sums sn = a1 + ⋯ + an

and Sn = b1 + ⋯ + bn .

Both are increasing sequences, and 0 ≤ sn ≤ Sn for each n. If the series ∑ bn is convergent, then by definition the sequence {Sn } is convergent, and hence bounded (Theorem 2.2). This implies that the increasing sequence {sn } is bounded as well. By the Monotone Convergence Theorem 2.3, the sequence {sn }, and hence the series ∑ an , is convergent. This proves the first assertion. The other assertion is the contrapositive of the first assertion.  Example 3.4. Applying the Comparison Test to the series ∑(2n2 + 5)−1 , we conclude from the convergence of the p-series ∑ n−2 and 0 < (2n2 + 5)−1 < n−2 that ∑(2n2 + 5)−1 is convergent. There are two common pitfalls when using the Comparison Test. First, the condition 0 ≤ an for all n, or at least all n ≥ N for some fixed N , cannot be omitted, as the following example illustrates. Example 3.5. Consider the case when an = −1 < n−2 = bn . Then ∑ bn is convergent, but ∑ an is divergent. The Comparison Test does not apply here because the condition, 0 ≤ an for all n, is not satisfied. Another common misconception of the Comparison Test is that the convergence of ∑ an implies that of ∑ bn . This is false, as the following example illustrates. Example 3.6. Consider the case when 0 < an = n−2 ≤ 1 = bn . Then ∑ an is convergent, but ∑ bn is divergent. The Comparison Test does not apply here because the hypotheses for both cases of the Comparison Test are not satisfied. 3.2.2

Limit Comparison Test

To motivate the second Comparison Test, consider the series 4 6 2n =2+ + + ⋯. 2 16 41 n=1 5n − 4 ∞

∑

2 When n is large, the quotient 5n2n 2 −4 is roughly equal to 5n . So the convergence 2 behaviors of the two series ∑ 5n2n 2 −4 and ∑ 5n should be the same. Since the harmonic 2 1 . Thus, we expect the series ∑ 5n2n series ∑ n is divergent (Example 3.1), so is ∑ 5n 2 −4 to be divergent as well. This is true, as the following test shows.

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Theorem 3.4 (Limit Comparison Test). Suppose that an = L ∈ R. an , bn > 0 for all n and lim n→∞ bn (1) If L > 0, then ∑ an is convergent if and only if ∑ bn is convergent. (2) If L = 0 and ∑ bn is convergent, then ∑ an is convergent. Ideas of Proof. The plan is to show that we can actually use the Comparison Test (Theorem 3.3). Proof.

If lim abnn = L > 0, then there exists a positive integer N such that n≥N

implies

L an 3L < < , 2 bn 2

which is equivalent to 3L L bn < an < bn . (3.4) 2 2 If ∑ an is convergent, then so is L2 ∑ an . So the left inequality in (3.4) and the Comparison Test imply that ∑ bn is convergent. Conversely, if ∑ bn is convergent, then so is 3L b . So the right inequality in (3.4) and the Comparison Test imply 2 ∑ n that ∑ an is convergent. The proof of the other assertion is an exercise.  2n Example 3.7. For the series ∑∞ n=1 an with an = 5n2 −4 , we use the Limit Comparison Test with bn = n1 . Then abnn → 25 > 0, so the divergence of the harmonic series ∑ n1 implies the divergence of the series ∑ 5n2n 2 −4 .

3.2.3

Exercises

(1) Prove Theorem 3.4 when L = 0. (2) Test the following series for convergence. (a) (b) (c) (d) (e) (f)

∑(3n2 − n + 4)−1 . ∑(4n − 3)−1/2 . √ ∑(2n + 3√n)−1 . ∑(2n + 3 n)−2 . ∑(n!)−1 . ∑(n2n )−1 .

(3) Suppose that an , bn > 0 for each n and that abnn → ∞. If ∑ bn is divergent, prove that ∑ an is divergent. (4) If ∑ an is convergent and each an ≥ 0, prove that ∑ a2n is convergent. k i j (5) Let p(x) = ∑m / 0. Prove i=0 ci x and q(x) = ∑j=0 dj x be polynomials with cm , dk = is convergent if and only if k ≥ m + 2. Here we assume that the series ∑ p(n) q(n) ) with N sufficiently large so that q(m) =/ 0 that the series starts at a term p(N q(N ) for all m ≥ N . This is known as the Polynomial Test.

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A First Course in Analysis

(6) Suppose that an , bn ≥ 0 for all n. (a) Prove that √ an bn ≤

an + bn 2

for all n. √ (b) Use the previous part to prove that ∑ an bn is convergent if both ∑ an and ∑ bn are convergent.

3.3

Alternating Series Test

In this section we discuss a convergence test for series whose terms have alternating signs. Definition 3.2. An alternating series is a series ∑(−1)n+1 an or ∑(−1)n an with each an > 0. Example 3.8. The alternating harmonic series ∞ n+1

∑ (−1) n=1

1 1 1 1 =1− + − +⋯ n 2 3 4

considered in Example 2.18 is an alternating series. There we showed that this series is convergent, in strong contrast with the harmonic series ∑ n1 (Example 3.1). The crucial ingredient in Example 2.18 is that the sequence { n1 } is decreasing. Of course, it is also true that n1 → 0. This example illustrates the ideas behind the following test. Theorem 3.5 (Leibnitz Alternating Series Test). Suppose that ● {an } is decreasing with each an > 0, and ● lim an = 0. Then the alternating series ∑(−1)n+1 an is convergent. Ideas of Proof. The plan is to recycle the argument in Example 2.18. We will show that the alternating series ∑(−1)n+1 an satisfies the Cauchy Convergence Criterion (Theorem 3.1). Proof. that

Let  > 0 be given. Since an → 0, there exists a positive integer N such

m≥N

implies am < .

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Then for n > m ≥ N we have n

∣ ∑ (−1)k+1 ak ∣ = ∣(−1)m+2 am+1 + (−1)m+3 am+2 + ⋯ + (−1)n+1 an ∣ k=m+1

= ∣am+1 − am+2 + am+3 − ⋯ + (−1)m+n+1 an ∣ = ∣am+1 − (am+2 − am+3 ) − (am+4 − am+5 ) −⋯∣ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ non-negative

non-negative

≤ am+1 < . This shows that the series ∑(−1)n+1 an satisfies the Cauchy Convergence Criterion and is, therefore, convergent.  Example 3.9. Consider the alternating series n+1 2

∑(−1)

1 n

n

. 1

1

We have that n−1 2 n > 0 and that the sequence {n−1 2 n } is decreasing. Moreover, 1 1 we have n−1 → 0 and 1 < 2 n ≤ 2 for all n, so n−1 2 n → 0. Thus, the hypotheses of the Alternating Series Test are all satisfied, and we conclude that the alternating series is convergent. One common mistake that students make about the Alternating Series Test is to use it to show that an alternating series is divergent. The Alternating Series Test cannot be used to show that any series is divergent. If one of its hypotheses is not satisfied, then the test simply does not apply. In this case, the series may converge or diverge. Example 3.10. The alternating series 2n 3n − 1 2n 2n → 32 , and so {(−1)n 3n−1 } does not converge to 0 (Corollary is divergent because 3n−1 3.2). The Alternating Series Test does not apply to this alternating series. The Alternating Series Test also does not apply to the alternating series 1 1 1 1 n+1 ∑(−1) an = 1 − 2 + 2 − 2 + 2 − ⋯ 3 2 5 4 because {an } is not decreasing. However, this alternating series is convergent by Exercises (11) and (17) on page 66 and Example 2.19. n ∑(−1)

3.3.1

Exercises

(1) Test the following series for convergence. (a) ∑(−1)n+1 (log n)−1 . (b) ∑(−1)n+1 n2 (n!)−1 . (c) ∑(−1)n+1 n−2 .

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(d) ∑(−1)n+1 n3 3−n . (2) Prove or disprove: If ∑ an is a convergent series and if bn → L, then ∑ an bn is convergent. (3) Prove or disprove: If ∑ an is convergent, then ∑ a2n is convergent. (4) Suppose that an , bn ≥ 0 for all n and that ∑ an and ∑ bn are both convergent. (a) Prove that ∑ an bn is convergent. (b) Is ∑ an bn still convergent if the hypothesis an , bn ≥ 0 for all n is omitted? (5) Suppose that ∑(−1)n+1 an is an alternating series that satisfies the hypotheses of the Alternating Series Test and has limit L. Prove that 0 < L < a1 . (6) Consider the alternating series ∑(−1)n+1 an in which an = n1 if n is odd and an = n12 if n is even. (a) Prove that an → 0. 1 for all odd n. (b) Prove that an − an+1 > n+1 (c) Prove that the alternating series ∑(−1)n+1 an is divergent. This series illustrates that the decreasing hypothesis in the Alternating Series Test cannot be omitted.

3.4

Absolute Convergence

Recall that the alternating harmonic series ∑(−1)n+1 n1 is convergent (Example 2.18), but the harmonic series ∑ n1 is divergent (Example 3.1). These two series tell us that, in general, ∑ an and ∑ ∣an ∣ can have very different convergence behaviors. To make precise the relationships between such series, we make the following definitions. 3.4.1

Absolute and Conditional Convergence

Definition 3.3. A series ∑ an is said to be ● absolutely convergent if the series ∑ ∣an ∣ is convergent. ● conditionally convergent if it is convergent, but ∑ ∣an ∣ is divergent. The alternating harmonic series ∑(−1)n+1 n1 is an example of a conditionally convergent series. So convergence does not imply absolute convergence. However, the other implication is true. Theorem 3.6. If a series ∑ an is absolutely convergent, then it is convergent. Proof. We will show that the series ∑ an satisfies the Cauchy Convergence Criterion (Theorem 3.1). Since ∑ an is absolutely convergent, given  > 0 there exists

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a positive integer N such that n > m ≥ N implies n

n

n

∣ ∑ ak ∣ ≤ ∑ ∣ak ∣ = ∣ ∑ ∣ak ∣∣ < . k=m+1

k=m+1

k=m+1

This shows that ∑ an is convergent.



In other words, absolute convergence implies convergence, but not vice versa. The following example illustrates that sometimes it is easier to show that a series is absolutely convergent than to show directly that it is convergent. . Some of its terms are negative because Example 3.11. Consider the series ∑ cos(n) n2 cos(n) can be negative for certain values of n. One would like to compare this series to the convergent p-series ∑ n12 with p = 2 (Example 3.3). However, neither one of the Comparison Tests applies because we cannot be sure that the nth term cos(n) n2 is positive. By Theorem 3.6 to show that ∑ cos(n) is convergent, it suffices to show n2 that it is absolutely convergent. Since ∣ cos(n)∣ ≤ 1 for any n, it follows that 0≤∣

1 cos(n) ∣ ≤ 2. 2 n n

Now the Comparison Test applies. Since the p-series ∑ n12 is convergent, so is cos(n) cos(n) ∑ ∣ n2 ∣. In other words, ∑ n2 is absolutely convergent, and hence convergent. The above example shows the usefulness of the concept of absolute convergence. Even if one is only interested in convergence, it is sometimes more convenient to consider absolute convergence. 3.4.2

Root Test

We now develop two commonly used tests for absolute convergence. Recall the concept of limit superior in section 2.4. Theorem 3.7 (Cauchy Root Test). Suppose that 1

L = lim sup ∣an ∣ n . Then the series ∑ an is ● absolutely convergent if L < 1, and ● divergent if L > 1. 1

Ideas of Proof. For large n, ∣an ∣ n is roughly equal to L, so ∣an ∣ is approximately Ln . Thus, the series ∑ ∣an ∣ behaves like the geometric series ∑ Ln , which is convergent if and only if ∣L∣ = L < 1. Proof. First suppose that L < 1. We need to show that ∑ ∣an ∣ is convergent. We can choose an  > 0 such that r = L +  < 1.

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will do. Since L is the largest subsequential limit For example, the choice  = 1−L 2 1 of the sequence {∣an ∣ n } (Theorems 2.12 and 2.13), this sequence is bounded. There are at most finitely many n such that 1

∣an ∣ n > L + . Thus, we can choose a positive integer N such that n≥N

1

implies ∣an ∣ n ≤ r < 1,

which in turn implies ∣an ∣ ≤ rn .

(3.5)

n

Since 0 < r < 1, the geometric series ∑ r is convergent. By (3.5) and the Comparison ∞ Test, the series ∑∞ n=N ∣an ∣, and hence ∑n=1 ∣an ∣, is convergent. Next suppose that L > 1. We can choose an  > 0 such that L −  > 1. There are infinitely many n such that 1

∣an ∣ n > L −  > 1, which implies that ∣an ∣ > 1 for infinitely many n. So {an } does not converge to 0, and the series ∑ an is divergent (Corollary 3.2).  n

3n ) . Then Example 3.12. Consider the series ∑ an with an = ( 5n−2 1

0 < ann =

3 3n → . 5n − 2 5

So in this case 3 <1 5 by Theorem 2.14. Thus, the Root Test says that the series ∑ an is absolutely convergent, and hence convergent. 1

lim sup ∣an ∣ n =

1

The Root Test gives no information if lim sup ∣an ∣ n = 1. In other words, when 1 lim sup ∣an ∣ n = 1, the series may converge or diverge (Exercise (3) below). 3.4.3

Ratio Test

The following test for absolute convergence will be used again when we discuss power series later in this book. Theorem 3.8 (d’Alembert Ratio Test). Suppose that an =/ 0 for all n. (1) If lim sup ∣ then ∑ an is absolutely convergent.

an+1 ∣ < 1, an

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(2) If lim inf ∣

an+1 ∣ > 1, an

then ∑ an is divergent. Ideas of Proof. Similar to the proof of the Root Test, the plan is to relate the series to a suitable geometric series, whose convergence behavior is known. Proof.

Suppose that

an+1 ∣ < 1. an To show that ∑ ∣an ∣ is convergent, first choose an  > 0 such that L = lim sup ∣

r = L +  < 1. There are at most finitely many n such that an+1 ∣ ∣ > r. an Thus, there exists a positive integer N such that an+1 n ≥ N implies ∣ ∣ ≤ r. an In particular, we have ∣aN +1 ∣ ≤ r∣aN ∣,

∣aN +2 ∣ ≤ r∣aN +1 ∣ ≤ r2 ∣aN ∣,

and so forth, leading to ∣aN +k ∣ ≤ rk ∣aN ∣

for all

k ≥ 0.

Therefore, we have k

k

∣aN ∣ 1 −r l=1 l=1 for all k ≥ 1, since 0 < r < 1. This shows that the sequence of partial sums of ∞ the series ∑∞ l=1 ∣aN +l ∣, which is increasing, is bounded as well. So ∑l=1 ∣aN +l ∣ is convergent by the Monotone Convergence Theorem 2.3. This implies that ∑ ∣an ∣ is convergent. The other assertion is left as an exercise for the reader.  l ∑ ∣aN +l ∣ ≤ ∑ r ∣aN ∣ <

Example 3.13. Consider the series ∑ an with an = 2nn . We have n + 1 2n n+1 1 an+1 ∣ = ∣ n+1 ⋅ ∣ = → < 1. ∣ an 2 n 2n 2 By the Ratio Test the series ∑ an is absolutely convergent, and hence convergent. n

Example 3.14. Consider the series ∑ an with an = xn! , where x is any real number. Then xn+1 n! ∣x∣ an+1 ∣=∣ ⋅ ∣= → 0 < 1, ∣ an (n + 1)! xn n+1 n

regardless of what x is. Therefore, by the Ratio Test the series ∑ xn! is absolutely convergent, and hence convergent, for all real numbers x.

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3.4.4

Exercises

(1) Test for convergence for the following series. (a) (b) (c) (d)

3

∑ 2nn . n . ∑ sin n2 − 12 ∑(n!) . ∑(log n)−n , n ≥ 2.

(2) In each case, determine the real numbers x such that the series converges. 2n

x . (a) ∑(−1)n (2n)!

(b) (c) (d) (e)

2n+1

x . ∑(−1)n (2n+1)! n+1 n nx . ∑(−1) xn . ∑ n ∑ n!xn . 1

(3) Denote lim sup ∣an ∣ n by L. (a) Give an example of a convergent series ∑ an with L = 1. (b) Give an example of a divergent series ∑ an with L = 1. Such examples illustrate that the Root Test is inconclusive when L = 1. (4) Finish the proof of Theorem 3.8. In other words, prove that if an =/ 0 for all n ∣ > 1, then ∑ an is divergent. and lim inf ∣ aan+1 n (5) Give examples to show that the Ratio Test is inconclusive when lim inf ∣

an+1 an+1 ∣ ≤ 1 ≤ lim sup ∣ ∣. an an

(6) Suppose that an > 0 for all n. and  > 0. Prove that there exists a positive integer N (a) Let L = lim sup aan+1 n such that aN an ≤ (L + )n (L + )N for all n ≥ N . (b) Use the previous part to prove that 1

lim sup (ann ) ≤ lim sup (

an+1 ). an

(c) Prove that 1 an+1 ) ≤ lim inf (ann ) . an This exercise illustrates that when the Ratio Test is applicable, then so is the Root Test. (7) Consider the series

lim inf (

1 1 3 1 2 3 1 2 3 2 1 3 3 2 + ( ) ( ) + ( ) ( ) + ( ) ( ) + ( ) ( ) + ⋯. 2 2 2 2 2 2 2 2 2

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= 12 and lim sup aan+1 = 23 . Thus, the Ratio Test is (a) Prove that lim inf aan+1 n n not applicable. √ 1 (b) Prove that ann → 23 . Thus, the Root Test shows that ∑ an is convergent. This example together with the previous exercise illustrate that the Root Test is more general than the Ratio Test. 1 (8) Suppose that an > 0 for all n and that lim sup ann < 1. Prove that ∑ np an is convergent for all positive integers p. 1 (9) Using Exercise (6b) or otherwise, prove that (n!) n → ∞. (10) Suppose that ∑ a2n and ∑ b2n are both convergent. a2 +b2

(a) Prove that ∣an bn ∣ ≤ n 2 n for each n. (b) Prove that ∑ an bn is absolutely convergent. (11) Suppose that an ≥ 0 for√ all n and that ∑ an is convergent. Using the previous a exercise, prove that ∑ n n is convergent. (12) Suppose that 1 1 ∣an ∣ ≤ − n n+1 for all n. Prove that ∑ an is absolutely convergent. (13) Consider the series 1 1 1 1 ∑ an = 1 + + + 2 + 2 + ⋯. 2 3 2 3 (a) Prove that the Ratio Test is not applicable. (b) Use the Root Test to determine if ∑ an is convergent. (14) Suppose that an > 0 for all n and that lim aan+1 = L > 0. n (a) Given  > 0 with  < L, prove that there exists a positive integer N such that aN aN (L − )n < an < (L + )n N (L − ) (L + )N for all integers n > N . 1 (b) Use the previous part to prove that ann → L. 3.5

Rearrangement of Series

Suppose that ∑ an is a convergent series. If the terms an are rearranged in the series, does the resulting series converge to the same limit? We will discuss this question in this section. Definition 3.4. If f ∶ Z+ → Z+ is a bijection, then ∑ af (n) = af (1) + af (2) + ⋯ is called a rearrangement of the series ∑ an .

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For a convergent sequence {an }, any rearrangement results in a sequence {af (n) } that also converges to the same limit (Exercise (8) on page 58). First we observe that this has an analog for absolutely convergent series. Theorem 3.9. Let ∑ an be an absolutely convergent series with ∑ ∣an ∣ = L, and let ∑ bn = ∑ af (n) be a rearrangement of ∑ an . Then ∑ bn is absolutely convergent with ∑ ∣bn ∣ = L. Ideas of Proof. The absolute convergence of ∑ bn will be established using the Monotone Convergence Theorem. Proof. First we show that ∑ ∣bn ∣ is convergent. Denote by sn and Sn the nth partial sums of the series ∑ ∣an ∣ and ∑ ∣bn ∣, respectively. Both {sn } and {Sn } are increasing. We have n

n

Sn = ∑ ∣bk ∣ = ∑ ∣af (k) ∣ ≤ sM ≤ L, k=1

(3.6)

k=1

where M = max{f (1), . . . , f (n)}. So {Sn } is both increasing and bounded. The Monotone Convergence Theorem 2.3 implies that {Sn }, and hence ∑ ∣bn ∣, is convergent. Denote its limit by L′ . We must show that L = L′ . Equation (3.6) implies that L′ = lim Sn ≤ L. On other hand, the series ∑ an is a rearrangement of ∑ bn , since an = bf −1 (n) . So the same argument as above shows that L ≤ L′ . Thus, we conclude that L = L′ .



Corollary 3.3. If ∑ an is a convergent series with each an ≥ 0, then any rearrangement ∑ af (n) of ∑ an also converges to the same limit. Proof.

When an ≥ 0 for all n, we have an = ∣an ∣,

and convergence of the series ∑ an is equivalent to absolute convergence. So we can apply Theorem 3.9.  The situation for conditionally convergent series is drastically different. Recall that a conditionally convergent series ∑ an is a convergent series that is not absolutely convergent. We now show that given a conditionally convergent series, every real number L is the limit of a rearrangement of the series.

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Theorem 3.10 (Riemann Rearrangement Theorem). Let ● ∑ an be a conditionally convergent series, and ● L be an arbitrary real number. Then there exists a rearrangement of ∑ an that converges to L. Ideas of Proof. The plan is to rearrange the positive and negative terms of the series such that the resulting partial sums sn oscillate above and below the intended limit L. The convergence of {sn } to L is guaranteed by the fact that the positive terms sequence and the negative terms sequence both converge to 0. Proof. Let pn and qn be the nth positive term and the nth negative term in {ak }, respectively. Since ∑ an is convergent, we have lim an = 0, which implies that lim pn = 0 = lim qn . Moreover, the conditional convergence of ∑ an implies that ∑ pn = ∞ and ∑ qn = −∞ by Exercise (2) on page 80. Now we construct the desired rearrangement. Take just enough, possibly zero, positive terms p1 , . . . , pn1 so that their sum exceeds L. In other words, we have p1 + ⋯ + pn1 −1 ≤ L < p1 + ⋯ + pn1 = sn1 . This is possible because ∑ pn = ∞. Now add just enough negative terms to the partial sum sn1 so that the sum is less than L, so sn1 +n2 = sn1 + q1 + ⋯ + qn2 < L ≤ sn1 + q1 + ⋯ + qn2 −1 . This is possible because ∑ qn = −∞. This process is now repeated ad infinitum. Add just enough positive terms pn1 +1 , . . . , pn1 +n3 to sn1 +n2 such that the sum sn1 +n2 +n3 exceeds L. Then add just enough negative terms to sn1 +n2 +n3 such that the sum is less than L, and so forth. Let ∑ af (n) be the rearrangement of ∑ an constructed in the previous paragraph, and let {sn } be its sequence of partial sums. To show that {sn } converges to L, let  > 0 be given. Since lim pn = 0 = lim qn , there exists a positive integer N0 such that   and ∣qn ∣ < . n ≥ N0 implies pn < 2 2 N0 N0 Let N be a sufficiently large integer such that {an }N n=1 includes {pn }n=1 and {qn }n=1 . Then for n ≥ N we have  ∣sn − L∣ ≤ sup{pk , ∣qk ∣ ∶ k ≥ N0 } ≤ < . 2 This proves that lim sn = L.  The proof above can be modified such that L can be taken to be the extended real numbers ±∞. Another modification of the proof above gives a rearrangement that is divergent but does not diverge to ±∞ (Exercise (3) below).

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3.5.1

Exercises

(1) Suppose that ∑ an converges absolutely and that ∑ bn is a rearrangement of ∑ an . If ∑ an = L, prove that ∑ bn = L. (2) Let ∑ an be a conditionally convergent series. Suppose that pn is the nth positive term in {ak } and that qn is the nth negative term in {ak }. (a) Prove that there are infinitely many pn and infinitely many qn . (b) Prove that pn → 0 and that qn → 0. (c) Prove that ∑ pn = ∞ and ∑ qn = −∞. (3) Suppose that ∑ an is a conditionally convergent series. (a) Prove that there exists a rearrangement of ∑ an that diverges to ∞. (b) Prove that there exists a rearrangement of ∑ an that diverges to −∞. (c) Prove that there exists a rearrangement of ∑ an that is divergent but does not diverge to ±∞. (4) Suppose that an ≥ 0 for all n and that ∑ an = ∞. Prove that ∑ af (n) = ∞ for every rearrangement ∑ af (n) of ∑ an . (5) Let ∑ an be a conditionally convergent series. Suppose that A < B, where A and B are extended real numbers. Prove that there exists a rearrangement ∑ af (n) such that lim inf sn = A and

lim sup sn = B,

where {sn } is the sequence of partial sums of ∑ af (n) . 3.6

Additional Exercises

(1) Suppose that {an } is a decreasing sequence with an > 0 for all n. If ∑ an is convergent, prove that nan → 0. This is known as Abel’s Theorem. Compare Abel’s Theorem with Corollary 3.2. n is convergent. (2) Prove that ∑ log n2 x n (3) Prove that ∑ ( n ) is convergent for every real number x. (4) Suppose that ∣an ∣ ≤ bn − bn+1 for all n, where {bn } is decreasing with bn → 0. Prove that ∑ an is absolutely convergent. (5) In each case, determine if the series is convergent or divergent. (a) 1 + (b) 1 + (c) 1 −

1 4 1 2 1 2

+ − −

1 7 1 3 1 3

+ + +

1 1 + 13 +⋯ 10 1 1 1 + − +⋯ 4 5 6 1 1 1 + 5 − 6 − 17 + ⋯ 4

(6) Suppose that an ≥ 0 for all n an (a) If ∑ an is convergent, prove that ∑ 1+a is convergent. n

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an is divergent. (b) If ∑ an is divergent, prove that ∑ 1+a n

(7) Suppose that ∑ an is absolutely convergent and that {bn } is bounded. Prove that ∑ an bn is absolutely convergent. (8) Prove or disprove: If ∑ an is convergent and if {bn } is bounded, then ∑ an bn is convergent. (9) Suppose that an → 0. Define bn = a2n−1 + a2n . Suppose that ∑ bn is convergent. (a) Prove that ∑ an is convergent. (b) Show by an example that ∑ an can be divergent if the hypothesis an → 0 is omitted. (10) Suppose that an → 0. Prove that there exists a subsequence ank such that ∞ ∑k=1 ank is absolutely convergent. (11) Suppose that an , bn > 0 for all n and that there exist a positive integer N and a real number r > 0 such that n≥N

implies an bn − an+1 bn+1 ≥ ran > 0.

(a) Prove that the sequence {an bn }∞ n=N is strictly decreasing. (b) Prove that the whole sequence {an bn } is convergent. (c) Prove that ∑∞ n=N (an bn −an+1 bn+1 ) is convergent with limit aN bN −L, where L = lim an bn . (d) Conclude that ∑ an is convergent. This is known as Kummer’s Test. (12) Let the setting be the same as in the previous exercise, and set an+1 . Kn = bn − bn+1 an (a) Suppose that K = lim inf Kn > 0 and that  > 0 with K −  > 0. Prove that there exists a positive integer N such that n≥N

implies Kn ≥ K − .

(b) If K = lim inf Kn > 0, prove that ∑ an is convergent. (c) Take bn = n − 1 and set Rn = Kn + 1. Prove that an+1 Rn = n (1 − ). an (d) If lim inf Rn > 1, prove that ∑ an is convergent. This is known as Raabe’s Test. (13) Using Raabe’s Test prove that the series x(x + 1) x(x + 1)(x + 2) x(x + 1)(x + 2)(x + 3) + + +⋯ 1+x+ 2! 3! 4! is convergent if x < 0. ∞ (14) Suppose that an , bn > 0 and that ∑∞ n=1 an and ∑n=1 bn are both convergent. For each n define n

cn = ∑ ai bn+1−i = a1 bn + a2 bn−1 + ⋯ + an−1 b2 + an b1 i=1

Denote the nth partial sums of ∑ an , ∑ bn , and ∑ cn by An , Bn , and Cn , respectively.

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(a) Prove that Cn ≤ An Bn for all n. (b) Prove that, for every integer n ≥ 1, there exists an integer N ≥ n such that An Bn ≤ CN . (c) Prove that ∑ cn is convergent with limit (∑ an )(∑ bn ). The series ∑ cn is called the Cauchy product of ∑ an and ∑ bn . (15) Let {an } and {bn } be two sequences, and let sn be the nth partial sum of ∑ bn . (a) For n > m ≥ 1, prove Abel’s Formula: n

n−1

∑ ak bk = an sn − am+1 sm + ∑ (ak − ak+1 )sk . k=m+1

k=m+1

(b) Suppose that {an } is decreasing with an → 0 and that {sn } is bounded. Prove that ∑ an bn is convergent. This is known as Dirichlet’s Test. (16) Using Dirichlet’s Test give another proof of the Alternating Series Test. (17) Suppose that {an } is a monotone convergent sequence and that ∑ bn is convergent. Using Dirichlet’s Test prove that ∑ an bn is convergent. This is known as Abel’s Test. Discuss why the monotone assumption on {an } is necessary.

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Chapter 4

Continuous Functions

In this chapter, we discuss continuous real valued functions defined on subsets of R. Continuity is one of the most basic and important concepts about real valued functions. As we will discuss later in this book, wherever a function is differentiable, it is continuous as well. Also, continuous functions provide a large class of integrable functions. Continuity is closely related to limits of a function, which we discuss in sections 4.1 and 4.2. Continuous functions are defined and discussed in section 4.3. In section 4.4 it is observed that when a continuous function f is defined on a closed bounded interval [a, b], it attains both a maximum and a minimum on that interval. It also satisfies the intermediate value property. In section 4.5 we discuss a concept called uniform continuity that is stronger than continuity. A function that is continuous on a closed bounded interval is automatically uniformly continuous. In general, however, a continuous function is not necessarily uniformly continuous. In section 4.6 we discuss monotone functions, the functional analogs of monotone sequences. It is shown that a monotone function on a bounded interval can have at most countably many points of discontinuity. If a strictly monotone function defined on a closed bounded interval is continuous, then so is its inverse function. In section 4.7 we discuss functions of bounded variation. Intuitively, a function of bounded variation is a function whose graph does not fluctuate too much. A function of bounded variation can be characterized as the difference of two increasing functions.

4.1

Limit Points

The limit of a function f at a point c is about the values of f (x) when x is close, but not equal, to c. This leads naturally to the concept of limit points. Definition 4.1. Let A be a non-empty subset of R, and let x be a real number. Then we say that x is a limit point of A if for every  > 0, there exist infinitely many elements a ∈ A such that 83

0 < ∣a − x∣ < .

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So x is not a limit point of A if and only if there exists an 0 > 0 such that there are only finitely many elements a ∈ A satisfying 0 < ∣a − x∣ < 0 . Note that a limit point of A is not required to be an element in A. Conversely, an element in A is not necessarily a limit point of A. Intuitively, if x is a limit point of A, then there are enough points in A that are close, but not equal, to x. In the following result, we provide two alternative characterizations of limit points. Theorem 4.1. Let A be a non-empty subset of R, and let x be a real number. The following statements are equivalent. (1) The number x is a limit point of A. (2) For every  > 0, there exists an element a ∈ A such that 0 < ∣a − x∣ < . (3) There exists a sequence {an } ⊆ A ∖ {x} such that lim an = x. Proof. (1) ⇒ (2) is immediate from the definition. To prove (2) ⇒ (3), by the hypothesis of (2), there exists an element a1 ∈ A such that 0 < ∣a1 − x∣ < 1. Since ∣a1 − x∣ > 0, there exists an element a2 ∈ A such that 1 1 0 < ∣a2 − x∣ < min {∣a1 − x∣, } ≤ . 2 2 Continuing this way we obtain a sequence {an } of elements in A such that 1 1 0 < ∣an − x∣ < min {∣an−1 − x∣, } ≤ n n for n ≥ 2. Since n1 → 0, we conclude that an → x. To prove (3) ⇒ (1), pick  > 0. Since we are now assuming that lim an = x, there exists a positive integer N such that n ≥ N implies ∣an − x∣ < . There are infinitely many an with n ≥ N , and ∣an − x∣ > 0 by the hypothesis of (3). So x is a limit point of A.  Corollary 4.1. Let A be a non-empty subset of R. Then a real number x is not a limit point of A if and only if there exists an 0 > 0 such that a ∈ A ∖ {x} implies ∣a − x∣ ≥ 0 . Proof.

This is the negation of condition (2) in Theorem 4.1.



{ n1

Example 4.1. The set A = ∶ n ∈ Z+ } has 0 ∈/ A as its only limit point. In fact, 1 since n → 0, it follows that 0 is a limit point of A. On other other hand, no real number x =/ 0 is a limit point of A by Corollary 4.1. Example 4.2. The set Q of rational numbers Indeed, if x is an arbitrary real number and  number a such that x < a < x +  (Theorem 1.3). which, by condition (2) in Theorem 4.1, implies

has R as its set of limit points. > 0, then there exists a rational This is equivalent to 0 < a − x < , that x is a limit point of Q.

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Exercises

(1) Prove that a non-empty finite set has no limit points. (2) Prove that every point c in an interval I is a limit point of I. (3) Suppose that a < b. Prove that the set of limit points of the open interval (a, b) is the closed interval [a, b]. (4) Prove that a closed interval is equal to its set of limit points. (5) Prove that c is a limit point of a non-empty subset A in R if and only if at least one of the following conditions is satisfied: (a) c is a limit point of {a ∈ A∶ a < c}. (b) c is a limit point of {a ∈ A∶ a > c}.

4.2

Limits of Functions

The purpose of this section is to discuss limits of a function. From now on, unless otherwise specified, whenever we say function, we mean a function whose domain Dom(f ) is a subset of R and whose target is R. 4.2.1

Limits

Definition 4.2. Let f ∶ A → R be a function, and let c be a limit point of A. Suppose that L is a real number. We say that the limit of f at c is L if for every sequence {an } in A ∖ {c}, lim an = c implies lim f (an ) = L. In this case, we write lim f = L or f → L as x → c, x→c

and say that f converges to L as x approaches c. If no such L exists, then we say that the limit of f at c does not exist. In particular, the limit of f at c is not L if there exists a sequence {an } in A∖{c} with lim an = c such that {f (an )} does not converge to L. From now on, whenever we use the symbol limx→c f , the point c is assumed to be a limit point of the domain of the function f . Observe that the limit of f at a point is really about the convergence of certain sequences {f (an )}, which we discussed in the previous two chapters. Many results about sequences can be translated into the context of limits of functions, some of which are in the exercises. The reader should be careful that in order to have limx→c f = L, the convergence f (an ) → L must hold for every sequence in A ∖ {c} with an → c. It is not enough to check f (an ) → L for just one sequence {an }. Be careful that limx→c f , whether it exists or not, is not about the value of f (c). In particular, limx→c f may exist even though f is not defined at c. Even if f is defined at c, the limit limx→c f is not necessarily equal to f (c).

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−x−1 with domain R ∖ { 21 }. We Example 4.3. Consider the function f (x) = 6x2x−1 want to compute limx→ 12 f , if it exists. Pick any sequence {an } in R ∖ { 21 } that converges to 21 . Since an =/ 21 for any n, we have

f (an ) =

1 5 6a2n − an − 1 = 3an + 1 → 3 ( ) + 1 = . 2an − 1 2 2

This shows that limx→ 21 f = 52 , even though f is not defined at 12 . Example 4.4. Consider the function f ∶ R → R defined as ⎧ 1 ⎪ ⎪sin ( x ) if x =/ 0, f (x) = ⎨ ⎪ if x = 0. ⎪ ⎩0 The reader should sketch the graph of f to visualize its behavior when x is close to 0. On any open interval containing 0, the graph of f fluctuates between −1 and 1 infinitely often. From its graph, one can guess that limx→0 f does not exist. To prove it, consider the sequence with an = (n−11 )π . We have an → 0, and 2

⎧ ⎪1 1 ⎪ f (an ) = sin (n − ) π = ⎨ ⎪ 2 ⎪ ⎩−1

if n is odd, if n is even.

So the sequence {f (an )} is divergent, showing that limx→0 f does not exist. 4.2.2

Uniqueness of Limits

As one can expect from the uniqueness of limits of sequences, the limit of a function at a point, if it exists, is unique. Theorem 4.2. Let f ∶ A → R be a function, and let c be a limit point of A. If the limit of f at c exists, then it is unique. Proof. Suppose that limx→c f = L and limx→c f = L′ . We want to show that L = L′ . Pick any sequence {an } in A ∖ {c} with lim an = c. Such a sequence exists by Theorem 4.1. Then lim f (an ) = L and

lim f (an ) = L′ .

Since the limit of a convergent sequence is unique (Theorem 2.1), we conclude that L = L′ .  4.2.3

-δ Characterization

Next we provide an alternative characterization of limits of a function in a form that is closer to the definition of uniform continuity, which will be discussed in the next section.

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Theorem 4.3. Let f ∶ A → R be a function, c be a limit point of A, and L be a real number. Then lim f = L

x→c

if and only if for every  > 0, there exists δ > 0 such that 0 < ∣x − c∣ < δ with x ∈ A

implies

∣f (x) − L∣ < .

Proof. For the “only if” part, suppose that limx→c f = L, and let  > 0 be given. We prove the existence of the required δ > 0 by contradiction, so assume that no δ > 0 satisfies the stated condition. For δ1 = 1, there must be an a1 ∈ A ∖ {c} with ∣a1 − c∣ < 1 Next, for δ2 =

min{ 12 , ∣a1

and

∣f (a1 ) − L∣ ≥ .

− c∣} > 0, there must be an a2 ∈ A ∖ {c} with ∣a2 − c∣ < δ2

and ∣f (a2 ) − L∣ ≥ .

Continuing this way we obtain a sequence {an } in A ∖ {c} with 1 and ∣f (an ) − L∣ ≥  ∣an − c∣ < n for each n. Since an → c and {f (an )} does not converge to L, we conclude that L is not the limit of f at c, which is a contradiction. This proves the “only if” part. For the “if” part, suppose that the stated -δ condition is satisfied. Suppose that {an } is an arbitrary sequence in A ∖ {c} with lim an = c. To show that f (an ) → L, let  > 0 be given. Using the δ > 0 from the -δ condition, we know that there exists a positive integer N such that n≥N

implies

0 < ∣an − c∣ < δ.

The -δ condition now implies that ∣f (an ) − L∣ <  for n ≥ N . This shows that limx→c f = L.



The -δ condition can be understood as follows. Given an  > 0, the value of f (x) can be made -close to L, provided that x ∈ A ∖ {c} is chosen to be δ-close to c. Be careful that first  > 0 is given, and then δ > 0 is chosen to make certain inequality true. The value of δ, in general, depends on both c and . Example 4.5. Consider the function f ∶ R → R defined as ⎧ 1 ⎪ ⎪x2 sin ( x ) if x =/ 0, f (x) = ⎨ ⎪2 if x = 0. ⎪ ⎩ From the graph of f , one can see that limx→0 f seems to be 0. We will prove this using the -δ characterization of limits. For x =/ 0 we have 1 ∣f (x) − 0∣ = ∣x2 sin ( )∣ ≤ x2 , x √ since ∣ sin(x)∣ ≤ 1 for all x. Given  > 0 we take δ =  > 0. Then 0 < ∣x − 0∣ < δ

implies

∣f (x) − 0∣ ≤ x2 < δ 2 = .

This shows that limx→0 f = 0, which is different from f (0).

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Recall that a convergent sequence is bounded (Theorem 2.2). If the limit of a function f exists at a point c, then it makes sense that f should be bounded near c. This is made precise in the following result. Theorem 4.4. Let f ∶ A → R be a function, and let c be a limit point of A. Suppose that limx→c f exists. Then there exist real numbers δ > 0 and M > 0 such that ∣x − c∣ < δ with x ∈ A Proof. that

implies

∣f (x)∣ ≤ M.

Say limx→c f is L. By Theorem 4.3, given  = 1 there exists δ > 0 such 0 < ∣x − c∣ < δ with x ∈ A implies ∣f (x) − L∣ < 1.

This in turn implies ∣f (x)∣ < ∣L∣ + 1. We now take ⎧ ⎪ ⎪max{∣f (c)∣, ∣L∣ + 1} if c ∈ A, M =⎨ ⎪ if c ∈/ A. ⎪ ⎩∣L∣ + 1 Then we have ∣f (x)∣ ≤ M whenever ∣x − c∣ < δ with x ∈ A. 4.2.4



One-Sided Limits

For some purposes, such as the discussion of monotone functions in section 4.6, it is convenient to consider limits when x approaches a point c from below or above. First we define limit from below. Definition 4.3 (Left-Hand Limits). Let f ∶ A → R be a function, c be a limit point of A
implies

lim f (an ) = L.

In this case, we write limx→c− f = L, and say that limx→c− f exists. (2) We say that the left-hand limit of f at c is ∞ (or −∞) if for every sequence {an } in A
implies

lim f (an ) = ∞

In this case, we write limx→c− f = ∞ (or −∞). The definition of limit from above is similar.

(or −∞).

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Definition 4.4 (Right-Hand Limits). Let f ∶ A → R be a function, c be a limit point of A>c = {x ∈ A ∶ x > c}, and L be a real number. (1) We say that the right-hand limit of f at c is L if for every sequence {an } in A>c , lim an = c

implies

lim f (an ) = L.

In this case, we write limx→c+ f = L, and say that limx→c+ f exists. (2) We say that the right-hand limit of f at c is ∞ (or −∞) if for every sequence {an } in A>c , lim an = c

implies

lim f (an ) = ∞

(or −∞).

In this case, we write limx→c+ f = ∞ (or −∞). Left-hand limits and right-hand limits are both called one-sided limits. If the one-sided limits limx→c− f and limx→c+ f are considered, then it is assumed that c is a limit point of {x ∈ Dom(f ) ∶ x < c} and {x ∈ Dom(f ) ∶ x > c}, respectively. As one can expect, limx→c f = L if and only if the two one-sided limits are equal to L. Moreover, one-sided limits can be characterized in terms of variations of the -δ condition in Theorem 4.3. These statements are left as exercises for the reader. 4.2.5

Exercises

(1) Write down an -δ characterization of limx→c f =/ L. (2) Suppose that f, g∶ A → R are functions, limx→c f = L, limx→c g = M , and a ∈ R. (a) (b) (c) (d)

Prove that limx→c af = aL, where (af )(x) = af (x) for x ∈ A. Prove that limx→c (f + g) = L + M , where (f + g)(x) = f (x) + g(x) for x ∈ A. Prove that limx→c (f g) = LM , where (f g)(x) = f (x)g(x) for x ∈ A. Suppose, in addition, that g(x) =/ 0 for any x ∈ A and M =/ 0. Prove that (x) L , where ( fg )(x) = fg(x) for x ∈ A. limx→c fg = M

(3) Let p(x) be a polynomial. Prove that limx→c p = p(c). (4) Given a function f ∶ A → R, define ∣f ∣∶ A → R by ∣f ∣(x) = ∣f (x)∣ for x ∈ A. (a) If limx→c f = L, prove that limx→c ∣f ∣ = ∣L∣. (b) Give an example in which limx→c ∣f ∣ exists, but limx→c f does not exist. √ (5) Given a function f ∶ A → R satisfying f (x) ≥ 0 for all x ∈ A, define f ∶ A → R √ √ by f (x) = f (x) for x ∈ A. √ √ (a) If limx→c f = L, prove that limx→c f = L. (b) Is the converse of the previous part true? (6) Suppose that a ≤ f (x) ≤ b for all x ∈ Dom(f ) for some real numbers a and b. If limx→c f = L, prove that a ≤ L ≤ b.

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(7) Suppose that f, g, h∶ A → R are functions such that f (x) ≤ g(x) ≤ h(x) for all x ∈ A. If limx→c f = L = limx→c h, prove that limx→c g = L. (8) Use the -δ characterization of limits to prove the following statements. √ √ (a) limx→2 x = 2. (b) limx→1 (x2 + 1)−1 = 1/2. (c) limx→1 3x2x 2 +1 = 1/2. (9) Determine if limx→0 x sin( x12 ) exists or not. (10) Suppose that limx→c f > 0. Prove that there exists δ > 0 such that ∣x − c∣ < δ with x ∈ Dom(f ) ∖ {c} implies f (x) > 0. (11) Prove that limx→c f = L if and only if both limx→c− f = L and limx→c+ f = L (12) (a) Prove that limx→c− f = L if and only if for every  > 0, there exists δ > 0 such that 0 < c − x < δ implies ∣f (x) − L∣ < . (b) Prove that limx→c+ f = L if and only if for every  > 0, there exists δ > 0 such that 0 < x − c < δ implies ∣f (x) − L∣ < . (13) (a) Prove that limx→c− f = ∞ if and only if for every real number M > 0, there exists δ > 0 such that 0 < c − x < δ implies f (x) > M . (b) Prove that limx→c− f = −∞ if and only if for every real number M < 0, there exists δ > 0 such that 0 < c − x < δ implies f (x) < M . (c) Formulate and prove similar statements for limx→c+ f = ±∞. (14) In each case, give an example that has the stated properties. (a) (b) (c) (d)

4.3

Both limx→c− f and limx→c+ f exist, but they are not equal. limx→c− f exists, but limx→c+ f does not exist. limx→c− f = L ∈ R and limx→c+ f = ∞. limx→c− f = ∞ and limx→c+ f = −∞.

Continuity

In this section we study continuous functions. When a continuous function is defined on a closed bounded interval, we will show in section 4.4 that it attains both a maximum and a minimum on that interval. Moreover, it satisfies the Intermediate Value Theorem. In section 4.5, we will discuss uniform continuity, a condition that is stronger than continuity. We will see that a continuous function on a closed bounded interval is uniformly continuous. 4.3.1

Sequential Definition of Continuity

For a function f to be continuous at a point a ∈ Dom(f ), what it should mean is that f (x) can be made as close to f (a) as one wants, provided that x is sufficiently close to a. As in the case of limits of a function, there are two equivalent ways to express this concept, one in terms of sequences and the other in terms of -δ. We begin with the first one.

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Definition 4.5. Let f ∶ A → R be a function, and let a ∈ A. We say that f is continuous at a if for every sequence {an } in A, lim an = a

implies

lim f (an ) = f (a).

If f is not continuous at a, we say that f is discontinuous at a. If f is continuous at each point in a non-empty subset B ⊆ A, we say that f is continuous on B. We call f ∶ A → R a continuous function if f is continuous on A. In particular, a function f is not continuous at a if there exists a sequence {an } in A with lim an = a such that {f (an )} does not converge to f (a). Notice that whether f is continuous at a has everything to do with the value of f (a), unlike the case of limx→a f . In the above definition, it is not required that a be a limit point of A. If a is not a limit point of A, then f is automatically continuous at a (Exercise (7) on page 111). On the other hand, if a is a limit point of A, then a comparison with Definition 4.2 shows that f is continuous at a if and only if lim f = f (a).

x→a

(4.1)

The reader should write down a proof of this assertion. Example 4.6. A polynomial p(x) is continuous on all of R. Indeed, the domain of p is R, so every point a is a limit point of Dom(p). Moreover, we have limx→a p = p(a) for every a (Exercise (3) on page 89). One should not expect that functions in general have simple continuity behaviors. The next two examples illustrate that continuity can be a delicate issue. Example 4.7. Consider the function χQ ∶ R → R defined as ⎧ ⎪ ⎪1 if x ∈ Q, χQ (x) = ⎨ ⎪ ⎪ ⎩0 if x ∈/ Q.

(4.2)

This is called the characteristic function of Q. We claim that χQ is discontinuous at every point a ∈ R. Indeed, first suppose that a ∈ Q, so χQ (a) = √ 1. Pick any sequence of irrational numbers {an } converging to a, such as an = a + n2 . Then χQ (an ) = 0 → 0 =/ χQ (a), showing that χQ is discontinuous at any a ∈ Q. On the other hand, suppose that b ∈/ Q, so χQ (b) = 0. Pick any sequence of rational numbers {bn } converging to b. The existence of such a sequence is guaranteed by Theorem 1.3. Then χQ (bn ) = 1 → 1 =/ χQ (b), so χQ is discontinuous at any b ∈/ Q.

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Example 4.8. Consider the function f ∶ R → R defined as ⎧ ⎪ ⎪x if x ∈ Q, f (x) = ⎨ ⎪ ⎪ ⎩0 if x ∈/ Q. Note that this function is the product f = x ⋅ χQ . We claim that f is continuous at 0 and is discontinuous at any x =/ 0. Indeed, since ∣f (x) − f (0)∣ = ∣f (x)∣ ≤ ∣x∣, if an → 0, then f (an ) → 0 = f (0). This shows that f is continuous at 0. On the other hand, if x =/ 0 and x ∈ Q, then we pick a sequence {an } of irrational numbers with an → x. Then f (an ) = 0 → 0 =/ x = f (x). If x ∈/ Q, then we pick a sequence {bn } of rational numbers with bn → x. Then f (bn ) = bn → x =/ 0 = f (x). Thus, f is discontinuous at any x =/ 0. 4.3.2

-δ Characterization of Continuity

Since limits can be characterized with an -δ condition, the same can be expected of continuity. Theorem 4.5. Let f ∶ A → R be a function, and let a ∈ A. Then f is continuous at a if and only if for ever  > 0, there exists δ > 0 such that ∣x − a∣ < δ with x ∈ A

implies

∣f (x) − f (a)∣ < .

The proof of this theorem is almost identical to that of Theorem 4.3 and will be left as an exercise for the reader. Example 4.9. Consider the function f (x) = x−1 defined on (0, 1). We show that it is continuous on (0, 1) using the -δ characterization of continuity. Pick any point a ∈ (0, 1), and let  > 0 be given. We need to estimate ∣a − x∣ 1 1 . ∣f (x) − f (a)∣ = ∣ − ∣ = x a xa To make this less than , we need a suitable lower bound of x. If we take a2  a δ = min { , }, 2 2 then a ∣x − a∣ < δ with x ∈ (0, 1) implies x > . 2 For such x, we have ∣a − x∣ 2∣a − x∣ 2δ ∣f (x) − f (a)∣ = < < 2 ≤ . xa a2 a By Theorem 4.5 this shows that f is continuous at a. Note that this δ is dependent on both a and . A different δ is needed if either the point a or  is changed.

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Exercises

(1) Prove Theorem 4.5. (2) Suppose a is a limit point of the domain of f . Prove that f is continuous at a if and only if (4.1) holds. (3) Suppose that f, g∶ A → R are continuous at a ∈ A. (a) Prove that cf is continuous at a, where c is any real number and (cf )(x) = cf (x) for x ∈ A. (b) Prove that f + g is continuous at a, where (f + g)(x) = f (x) + g(x) for x ∈ A. (c) Prove that f g is continuous at a, where (f g)(x) = f (x)g(x) for x ∈ A. (d) Suppose, in addition, that g(a) =/ 0. Prove that fg is continuous at a. (4) Suppose that f ∶ A → R is continuous at a. If ∣f ∣∶ A → R is defined as ∣f ∣(x) = ∣f (x)∣ for x ∈ A, prove that ∣f ∣ is continuous at a. Is the converse true? (5) Suppose that f ∶ A → R is continuous at a and f (x) ≥ 0 for all x ∈ A. If √ √ √ √ f ∶ A → R is defined as f (x) = f (x) for x ∈ A, prove that f is continuous at a. Is the converse true? (6) Suppose that f and g are functions with Ran(f ) ⊆ Dom(g). Define the composition of f and g by (g ○ f )(x) = g(f (x)) for x ∈ Dom(f ). If f is continuous at a and g is continuous at f (a), prove that g ○ f is continuous at a. (7) Suppose that f is continuous at a (a) If f (a) > 0, prove that there exists δ > 0 such that ∣x − a∣ < δ with x ∈ Dom(f ) implies f (x) > 0. (b) If f (a) < 0, prove that there exists δ > 0 such that ∣x − a∣ < δ with x ∈ Dom(f ) implies f (x) < 0. (8) For a subset S ⊆ R, define its characteristic function χS ∶ R → R by ⎧ ⎪ ⎪1 if x ∈ S, χS (x) = ⎨ ⎪ ⎪ ⎩0 if x ∈/ S. Determine the points of continuity of χZ , χI , and x ⋅ χI . 4.4

Extreme and Intermediate Value Theorems

One reason why continuous functions are important is that they have certain good properties when defined on an interval. The first such result says that a continuous function defined on a closed bounded interval is bounded and attains both a maximum and a minimum. To be precise, we need the following definition. Definition 4.6. A function f ∶ A → R is said to be bounded if there exists M >0 for all x ∈ A.

such that ∣f (x)∣ ≤ M

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Theorem 4.6 (Extreme Value Theorem). Let f ∶ [a, b] → R be a continuous function. Then f is bounded. Moreover, there exist α and β in [a, b] such that f (α) ≤ f (x) ≤ f (β) for all x ∈ [a, b]. Ideas of Proof. That f is bounded will be shown to be a consequence of the Bolzano-Weierstrass Theorem and the fact that a convergent sequence is bounded. The existence of α is shown by constructing a sequence in [a, b] whose image under f converges to the infimum of f on [a, b]. Proof. We first prove that f is bounded by contradiction. If f is not bounded, then there exists a sequence {an } in I = [a, b] such that ∣f (an )∣ → ∞. Since {an } is bounded, it has a convergent subsequence {ank } by the Bolzano-Weierstrass Theorem 2.8. We also have ∣f (ank )∣ → ∞. Since f is continuous and {ank } is convergent, the sequence {f (ank )} is convergent, and hence bounded (Theorem 2.2). This is a contradiction, so f is bounded. Next we show the existence of α ∈ I such that f (α) ≤ f (x) for all x ∈ I. We just proved in the previous paragraph that the set {f (x) ∶ x ∈ I} is bounded, so m = inf{f (x) ∶ x ∈ I} is a real number. It suffices to show that f (α) = m for some α ∈ I. For each n ∈ Z+ , there must exist xn ∈ I such that m ≤ f (xn ) < m +

1 , n

so we have f (xn ) → m. By the Bolzano-Weierstrass Theorem 2.8 again, {xn } has a convergent subsequence {xnk } with limit α ∈ I because I is a closed bounded interval. Since f is continuous at α, it follows that f (xnk ) → f (α). But {f (xnk )} is a subsequence of {f (xn )}, so f (xnk ) → m (Theorem 2.6). The uniqueness of the limit of a convergent sequence (Theorem 2.1) now implies that m = f (α). The existence of β is proved similarly by considering M = sup{f (x) ∶ x ∈ I} instead of m. The details are left to the reader as an exercise.  The reader should be careful that the hypothesis of I being a closed bounded interval cannot be omitted from the Extreme Value Theorem. The next result says that a continuous function f defined on a closed bounded interval [a, b] has the intermediate value property: If r is strictly between f (a) and f (b), then r = f (x) for some x ∈ [a, b]. This seems pretty obvious. However, it does take a bit of work to give a rigorous proof of this fact. Theorem 4.7 (Intermediate Value Theorem). Let f ∶ [a, b] → R be a continuous function such that f (a) =/ f (b). If r lies strictly between f (a) and f (b), then there exists a point x0 ∈ [a, b]

such that

r = f (x0 ).

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Ideas of Proof. We show the existence of x0 by considering the supremum of the set of points in [a, b] whose images under f is < r. Proof.

There are two possibilities. We have either f (a) < r < f (b) or

f (b) < r < f (a).

We will consider the first case and leave the similar second case to the reader as an exercise. Consider the set S = {x ∈ [a, b] ∶ f (x) < r}, which is non-empty because a ∈ S. Its least upper bound x0 = sup S is still in [a, b]. We will prove that f (x0 ) = r. For each n ∈ Z+ , there exists xn ∈ S such that x0 −

1 < xn ≤ x0 , n

so we have xn → x0 . By the continuity of f , we have f (xn ) → f (x0 ), so f (x0 ) ≤ r because each f (xn ) < r. It remains to show that f (x0 ) ≥ r as well. Consider wn = min {b, x0 +

1 } ∈ [x0 , b] ⊆ [a, b]. n

Each wn satisfies x0 ≤ wn ≤ x0 +

1 , n

so wn → x0 . The continuity of f implies that f (wn ) → f (x0 ). Now since x0 +

1 > x0 = sup S n

and f (b) > r,

we know that wn ∈/ S. Therefore, we have f (wn ) ≥ r for each n. This implies that f (x0 ) ≥ r, as desired.



It should be emphasized that the previous two theorems give sufficient conditions, namely, being continuous on a closed bounded interval, that guarantee that a function has the extreme and intermediate value properties. However, being continuous on a closed bounded interval is not necessary for a function to have such properties. For example, as we will see in the next chapter, the derivative of a function need not be continuous. However, it does have the intermediate value property.

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A First Course in Analysis

Exercises

(1) Finish the proof of Theorem 4.6 by proving the existence of β. (2) Finish the proof of Theorem 4.7 by proving the case f (b) < r < f (a). (3) Let f ∶ I → R be a continuous function, where I is a closed bounded interval. Prove that f (I) = {f (x) ∶ x ∈ I} is either a single point or a closed bounded interval. (4) Let f ∶ I → R be a continuous function, where I is an interval. Prove that f (I) is either a single point or an interval. (5) Let f ∶ [a, b] → R be a continuous function such that f (a)f (b) < 0. Prove that f (c) = 0 for some c ∈ (a, b). (6) Let f ∶ [0, 1] → [0, 1] be a continuous function. Prove that it has a fixed point, i.e., a point a ∈ [0, 1] such that f (a) = a. (7) Show by a counterexample that the conclusions of the Extreme Value Theorem do not have to hold if the domain of the continuous function is an open bounded interval. (8) Give an example of a function f ∶ [a, b] → R that is not continuous somewhere in [a, b] but that satisfies the conclusions of both the Extreme Value Theorem and the Intermediate Value Theorem.

4.5

Uniform Continuity

In the -δ characterization of continuity, it makes sense that smaller values of δ > 0 are needed if  > 0 is getting smaller. For a fixed  > 0, is it possible that one δ will work for all the points a in Dom(f )? We need the following concept to make this idea precise. Definition 4.7. A function f ∶ A → R is said to be uniformly continuous on A if for every  > 0, there exists δ > 0 such that ∣a0 − a1 ∣ < δ with a0 , a1 ∈ A implies ∣f (a0 ) − f (a1 )∣ < . So a function f ∶ A → R is not uniformly continuous on A if and only if there exists 0 > 0 such that for every δ > 0, there exist a0 , a1 ∈ A with ∣a0 − a1 ∣ < δ

such that ∣f (a0 ) − f (a1 )∣ ≥ 0 .

To explain the definition a bit further, a function f ∶ A → R is uniformly continuous if and only if the following is true. For each  > 0, there exists δ > 0 such that, given any two points a0 and a1 in A that are within δ of each other, their images under f are guaranteed to be within  of each other. It is in this sense that this δ works for all the points in A. We now show that uniform continuity implies continuity. Theorem 4.8. Let f ∶ A → R be uniformly continuous on A. Then f is continuous on A.

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Proof. Pick any point a ∈ A. To prove that f is continuous at a, let  > 0 be given. Since f is uniformly continuous on A, there exists δ > 0 such that ∣a0 − a1 ∣ < δ with a0 , a1 ∈ A implies ∣f (a0 ) − f (a1 )∣ < . In particular, we have that ∣x − a∣ < δ with x ∈ A implies ∣f (x) − f (a)∣ < , so f is continuous at a.



The converse of the above theorem is false. In other words, there are continuous functions that are not uniformly continuous. The next example gives one such function. Example 4.10. Consider once again the function f (x) = x−1 with domain (0, 1). We showed in Example 4.9 that f is continuous on (0, 1). Now we show that f is not uniformly continuous on (0, 1). With 0 = 1 we will show that no δ > 0 can satisfy the condition in Definition 4.7. Indeed, given any δ > 0, we can take a0 1 a0 = min {δ, } and a1 = . 2 2 Then a0 δ ∣a0 − a1 ∣ = ≤ < δ. 2 2 Moreover, we have 1 1 1 > 1 = 0 . ∣ − ∣= a0 a1 a0 So f is not uniformly continuous on (0, 1). Theorem 4.8 and Example 4.10 together imply that uniform continuity is strictly stronger than continuity in general. This leads naturally to the following question. Is there a simple additional condition that can guarantee that a continuous function is uniformly continuous? The following result provides one simple answer to this question. Theorem 4.9. Let f ∶ I → R be a continuous function, where I = [a, b] is a closed bounded interval. Then f is uniformly continuous on I. Ideas of Proof. We will show that, if f is not uniformly continuous, then there are two sequences whose images under f are separated by a certain fixed distance. But they also have subsequences converging to the same point. The continuity of f will then lead to a contradiction. Proof. We prove this by contradiction. If f is not uniformly continuous on I, then there exists 0 > 0 such that for every δ > 0, there exist a0 , a1 ∈ I with ∣a0 − a1 ∣ < δ

and ∣f (a0 ) − f (a1 )∣ ≥ 0 .

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So for each n ∈ Z+ , there exist xn , wn ∈ I such that ∣xn − wn ∣ <

1 n

and ∣f (xn ) − f (wn )∣ ≥ 0 .

(4.3)

Since the sequence {xn } is bounded, by the Bolzano-Weierstrass Theorem 2.8 it has a convergent subsequence {xnk } with limit, say, x ∈ I. The continuity of f implies that f (xnk ) → f (x). Moreover, we also have ∣xnk − wnk ∣ <

1 , nk

so wnk → x as well. The continuity of f then implies that f (wnk ) → f (x). Thus, the sequence {f (xnk ) − f (wnk )} converges to 0, which contradicts (4.3). So f must be uniformly continuous.  4.5.1

Exercises

(1) Let a > 0 be a real number. Consider the function f ∶ [a, ∞) → R defined as f (x) = x−1 . Prove that f is uniformly continuous. (2) Consider the function f (x) = x2 with domain R. Prove that f is not uniformly continuous. (3) Prove that the function f ∶ (0, 1) → R defined as f (x) = x−2 is not uniformly continuous. (4) Let f, g∶ A → R be uniformly continuous functions, and let c be a real number. (a) Prove that cf is uniformly continuous. (b) Prove that f + g is uniformly continuous. (c) Give an example in which f g is not uniformly continuous. In other words, the product of two uniformly continuous functions is not necessarily uniformly continuous. (5) Give an example in which neither f ∶ A → R nor g∶ A → R is uniformly continuous, but the product f g∶ A → R is uniformly continuous. (6) Let f ∶ A → R be uniformly continuous, and let {an } be a Cauchy sequence in A. (a) Prove that {f (an )} is a Cauchy sequence. (b) Show by an example that the previous part is false if f is only assumed to be continuous. (7) Let f ∶ (a, b) → R be uniformly continuous. Show that f can be extended to a continuous function on [a, b] as follows. (a) If {xn } is a convergent sequence in (a, b), prove that {f (xn )} is a convergent sequence. (b) Suppose that {xn } and {yn } are two sequences in (a, b) converging to a. Prove that the limits of {f (xn )} and {f (yn )} are equal.

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(c) If {xn } is any sequence in (a, b) converging to a, define f (a) = lim f (xn ). Prove that f (a) is well-defined and that the extended function f is continuous at a. (d) Repeat the steps above for the other end point b to extend f to a continuous function on [a, b]. The continuous function f ∶ [a, b] → R constructed above is called a continuous extension of the original function f . (8) Give an example of a bounded continuous function f ∶ (a, b) → R that cannot be extended to a function that is continuous at a.

4.6

Monotone and Inverse Functions 1

For an integer n ≥ 2, it is possible to show directly that the function g(x) = x n defined on [0, ∞) is continuous. However, whether one uses the sequential definition or the -δ characterization of continuity, showing that g is continuous directly does involve quite a bit of work, especially if n is large. For example, try it for n = 7. Notice that g is the inverse function of f (x) = xn defined on [0, ∞). Since h(x) = x is continuous on R, it follows that f is continuous as well (Exercise (3) on page 93). This leads naturally to the following question. Is it possible to conclude from the continuity of f that its inverse function is continuous? If this is true, then it would save us a lot of work when dealing with inverse functions. As you will see in this section, under some reasonable assumptions, the answer to the question above is yes. In particular, we can, in fact, conclude from the 1 continuity of f (x) = xn , with domain [0, ∞), that its inverse function g(x) = x n is continuous on [0, ∞). We begin with some relevant definitions. 4.6.1

Monotone Functions

Definition 4.8. Let f ∶ A → R be a function. (1) (2) (3) (4) (5) (6)

We say that f is increasing if a < b in A implies f (a) ≤ f (b). We say that f is strictly increasing if a < b in A implies f (a) < f (b). We say that f is decreasing if a < b in A implies f (a) ≥ f (b). We say that f is strictly decreasing if a < b in A implies f (a) > f (b). We say that f is monotone if f is either increasing or decreasing. We say that f is strictly monotone if f is either strictly increasing or strictly decreasing.

Monotone functions will appear again in the next section when we discuss functions of bounded variation. Moreover, they are often used in applications in statistics, physics, and engineering. For example, cumulative distribution functions in probability theory are monotone functions. Furthermore, monotone functions de-

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fined on a closed bounded interval are Riemann integrable, as we will show in a later chapter. 1

Example 4.11. For a positive integer n, both f (x) = xn and g(x) = x n with domain [0, ∞) are strictly increasing. The function x2 with domain [−1, 1] is neither increasing nor decreasing. A function h is (strictly) increasing if and only if −h is (strictly) decreasing. A strictly monotone function f is injective, so it has an inverse function f −1 ((1.1) on p. 6). We want to show that if f is strictly monotone and continuous on an interval, then its inverse function is also strictly monotone and continuous. To prove this, we need the following result. Theorem 4.10. Let f ∶ I → R be a strictly monotone function on an interval I such that f (I) is also an interval. Then f is continuous on I. Ideas of Proof. One can visualize this result as follows. If f is not continuous at some point a ∈ I, then since f is strictly monotone, there must be a “jump” in the graph of f at the point a. But this jump would violate the hypothesis that f (I) is an interval. So f must be continuous on I. Proof. We will consider the case when f is strictly increasing. The strictly decreasing case can be dealt with by considering −f . Pick a point a ∈ I. We must show that f is continuous at a. Let  > 0 be given. First consider the case when a is not an end point of I. Since a is not an end point of I and f is strictly increasing, f (a) is not an end point of the interval f (I). So there exists δ0 > 0 with δ0 ≤  such that the closed interval [f (a) − δ0 , f (a) + δ0 ] is contained in f (I). There exist unique elements α and β in I with α < a < β such that f (α) = f (a) − δ0

and f (β) = f (a) + δ0 .

If we take δ = min{a − α, β − a}, then ∣x − a∣ < δ

implies α ≤ a − δ < x < a + δ ≤ β.

Since f is strictly increasing, we have f (α) < f (x) < f (β). This implies that f (a) −  ≤ f (a) − δ0 = f (α) < f (x) < f (β) = f (a) + δ0 ≤ f (a) + , from which we conclude that ∣f (x) − f (a)∣ < . This proves that f is continuous at a. Next suppose a is the left end point of I, so I = [a, b], [a, b), or [a, ∞). Then f (a) is the left end point of f (I) because f is strictly increasing. There exists δ0 > 0 with δ0 ≤  such that [f (a), f (a) + δ0 ] ⊆ f (I).

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So there exists a unique point β ∈ I with β>a

and f (β) = f (a) + δ0 .

If ∣x − a∣ < δ = β − a with x ∈ I, then a≤x<β

and f (a) ≤ f (x) < f (β) ≤ f (a) + .

Thus, we have 0 ≤ f (x) − f (a) < , showing that f is continuous at a. A similar argument can be used when a is the right end point of I. The details of this case is left to the reader as an exercise.  4.6.2

Continuity of Inverse Functions

We now use the previous theorem to show that the inverse function of a strictly monotone continuous function on an interval is also continuous. Theorem 4.11. Let f be a strictly monotone continuous function whose domain is an interval I. Then its inverse function f −1 is a strictly monotone continuous function on f (I). Proof. Since f is strictly monotone, it is injective and has an inverse function f −1 with domain f (I) and range I. Since f is continuous and injective, the set f (I) is an interval (Exercise (4) on page 96). We first show that f −1 is also strictly monotone. Suppose that f is strictly increasing. We will show that f −1 is also strictly increasing. Pick y0 < y1 in f (I). Then there are unique points x0 and x1 in I such that f (x0 ) = y0

and f (x1 ) = y1 .

Since f is strictly increasing and y0 < y1 , we have x0 < x1 . By the definition of f −1 , we have f −1 (y0 ) = x0 < x1 = f −1 (y1 ). This shows that f −1 is also strictly increasing. A similar argument shows that if f is strictly decreasing, then so is f −1 . Now f −1 is strictly monotone, whose domain is the interval f (I). Moreover, we have f −1 (f (I)) = I, which is also an interval. Thus, Theorem 4.10 implies that f −1 is continuous on f (I).  Example 4.12. For every positive integer n, the function f (x) = xn defined on [0, ∞) is strictly increasing and continuous. Its range is also [0, ∞). Theorem 1 4.11 tells us that its inverse function f −1 (x) = x n is also strictly increasing and continuous on [0, ∞).

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Points of Discontinuity

While a monotone function defined on an interval may not be continuous at all the points of its domain, we now show that it does not have too many points of discontinuity. Theorem 4.12. Let f ∶ (a, b) → R be a monotone function for some a < b. Then f is discontinuous at at most countably many points in (a, b). Ideas of Proof. We show that, if f is not continuous at a point c, then the monotonicity of f forces it to have a “jump” at c, inside of which we will pick a rational number. The countability of Q will then be used to make the desired conclusion. Proof. We will consider the case when f is increasing. The decreasing case can be dealt with by considering −f . It suffices to show that the set of points of discontinuity of f is a subset of a countable set. Suppose that f is discontinuous at a point c in (a, b). Since c is a limit point of (a, b), we know that f (c) =/ lim f.

(4.4)

x→c

Since f is increasing, both αc = sup{f (x) ∶ c > x ∈ (a, b)} and βc = inf{f (x) ∶ c < x ∈ (a, b)} are real numbers. We claim that αc = lim− f

and βc = lim+ f.

x→c

x→c

To prove this, we use the -δ characterization of one-sided limits (Exercise (12) on page 90). Let  > 0 be given. There must exist x0 < c in (a, b) such that αc −  < f (x0 ) ≤ αc . Take δ = c − x0 . Then 0
implies x0 < x < c,

so we have αc −  < f (x0 ) ≤ f (x) ≤ αc . So ∣f (x) − αc ∣ < 

if

0 < c − x < δ,

and we conclude that limx→c− f = αc . A similar argument shows that limx→c+ f = βc . Now if both inequalities in lim f = αc ≤ f (c) ≤ βc = lim+ f

x→c−

x→c

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are equalities, then we would have f (c) = limx→c f (Exercise (11) on page 90), contradicting (4.4). Thus, one of the above inequalities must be strict, so αc < βc . Choose any rational number qc in (αc , βc ), which is possible by Theorem 1.3. Starting at a point c of discontinuity of f , we have associated a rational number qc to it. Moreover, if d > c is another point of discontinuity of f , then the increasing hypothesis on f implies that βc ≤ αd , so qc < βc ≤ αd < qd . Since distinct points of discontinuity of f give rise to distinct rational numbers and the set Q of rational numbers is countable, the theorem is proved.  An interval (a, b) is an uncountable set. Theorem 4.12 says that a monotone function defined on (a, b) is continuous except possibly on a countable or finite subset. There is a sort of converse to Theorem 4.12. Suppose that I is a bounded interval and that S is a finite or countable subset of I. Then there exists a monotone function f with domain I whose set D of points of discontinuity is exactly S. In fact, we can even insist that f be strictly increasing (or strictly decreasing, if one wishes). These issues are explored further in the exercises. 4.6.4

Exercises 1

(1) Consider the function g(x) = x n for some positive integer n. (a) Prove that g(x) is uniformly continuous on [0, 1]. (b) Prove that g(x) is uniformly continuous on [1, ∞). (c) Using the previous two parts or otherwise, prove that g(x) is uniformly continuous on [0, ∞). (2) For integers n ≥ 2, prove that f (x) = xn is not uniformly continuous on [0, ∞). Together with the previous exercise, this shows that uniform continuity is not preserved by the process of taking inverse functions. (3) Suppose that f ∶ I → R is continuous on an interval I = [a, b] and that f is injective. (a) If f (a) < f (b), prove that f is strictly increasing. (b) If f (a) > f (b), prove that f is strictly decreasing. (4) Suppose that f, g∶ A → R are increasing functions. (a) For a real number c, prove that cf is increasing if c > 0 and is decreasing if c < 0. (b) Prove that the sum f + g is increasing. (5) Suppose that f, g∶ A → R are increasing functions. (a) Give an example in which the product f g is not monotone. (b) If, in addition, f (x) > 0 and g(x) > 0 for all x ∈ A, prove that f g is increasing.

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(6) Suppose that there is a sequence of disjoint intervals In (n ∈ Z+ ) such that R = ⋃∞ n=1 In . Let f ∶ R → R be a function such that f is monotone on each In . Prove that f is continuous on R, except possibly on a finite or countable set. (7) Give an example of a bijective function f ∶ [0, 1] → [0, 1] whose inverse function f −1 is not continuous at at least one point in [0, 1]. (8) Let f ∶ [a, b] → R be a monotone function. Prove that the set of points of discontinuity of f is either finite or countable. (9) Let f ∶ (a, b) → R be an increasing function, and let c be a point in (a, b). (a) Prove that lim f − lim− f = inf{f (w) − f (x) ∶ x < c < w, x, w ∈ I}.

x→c+

x→c

(b) Prove that f is continuous at c if and only if inf{f (w) − f (x) ∶ x < c < w, x, w ∈ I} = 0. (10) Consider the step function J∶ R → R defined as ⎧ ⎪ ⎪0 J(x) = ⎨ ⎪ ⎪ ⎩1

if x < 0, if x ≥ 0.

(a) Prove that J is continuous on R ∖ {0} and discontinuous at 0. (b) Suppose that x1 < x2 < ⋯ < xn are n distinct points in R and a1 , . . . , an are n positive real numbers. Sketch the graph of the function n

f (x) = ∑ ak J(x − xk ).

(4.5)

k=1

(c) Prove that f is an increasing function. (d) Prove that f is continuous on R ∖ {x1 , . . . , xn } and discontinuous at x1 , . . . , x n . The function f is thus a monotone function with a prescribed finite set of points of discontinuity. (11) Suppose that x1 < x2 < ⋯ < xn are n distinct points in R. Construct a strictly increasing function f ∶ R → R that is continuous on R ∖ {x1 , . . . , xn } and discontinuous at x1 , . . . , xn . 4.7

Functions of Bounded Variation

We saw in the previous section that monotone functions have some nice properties. For example, a monotone function is continuous except possibly on a countable set. They are also closed under scalar multiplication, and the sum of two increasing functions is increasing (Exercise (4) on page 103). However, the difference and product of two monotone functions are not necessarily monotone (Exercise (5) on page 103).

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For many purposes it is convenient to have a larger class of functions, containing all the monotone functions, that are closed under taking scalar multiples, sums, and products. In particular, such a class of functions must contain functions of the form f = g − h, where g and h are increasing. The class of differences of increasing functions, called functions of bounded variation, has the desired properties (see the exercises). 4.7.1

Variation

We will define functions of bounded variation in terms of the concept of variation, which measures how much the function fluctuates. Then we will show that they are equivalent to differences g − h of increasing functions. For the definition below, we need the concept of partition. Definition 4.9. A partition of [a, b] is a finite ordered set P = {a = x0 < x1 < ⋯ < xn = b}. The points x0 , . . . , xn are called the partition points of P . So a partition P divides the interval [a, b] into n sub-intervals. Here n can be any positive integer, so the simplest partition of [a, b] is P = {a < b}. Definition 4.10. Let f ∶ [a, b] → R be a function, and let [c, d] ⊆ [a, b] be a subinterval. (1) Suppose P = {x0 < ⋯ < xn } is a partition of [c, d]. The variation of f with respect to P is the sum n

V (f ; P ) = ∑ ∣f (xk ) − f (xk−1 )∣. k=1

(2) The variation of f on [c, d] is defined as the extended real number V (f ; [c, d]) = sup {V (f ; P ) ∶ P a partition of [c, d]} . If V (f ; [c, d]) is finite, then we say that f is of bounded variation on [c, d]. Otherwise, we take V (f ; [c, d]) = ∞. The simplest functions of bounded variation are monotone functions. Theorem 4.13. Every monotone function on [a, b] is of bounded variation. Proof. Suppose f is an increasing function on [a, b]. Let P = {x0 < ⋯ < xn } be any partition of [a, b]. Then we have a telescoping sum: n

V (f ; P ) = ∑ ∣f (xk ) − f (xk−1 )∣ k=1

= (f (x1 ) − f (x0 )) + (f (x2 ) − f (x1 )) + ⋯ + (f (xn ) − f (xn−1 )) = f (xn ) − f (x0 ) = f (b) − f (a) < ∞.

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Since this is true for any partition P of [a, b], we have shown that V (f ; [a, b]) = f (b) − f (a) for an increasing function f , which is, therefore, of bounded variation. The case where f is decreasing is left as an exercise.  There are, however, functions of bounded variation that are not monotone. Example 4.13. Consider the function f (x) = ∣x∣ defined on [−1, 1], which is not monotone. We show that f is of bounded variation. With the partition {−1, 0, 1}, we see that V (f ; [−1, 1]) ≥ ∣f (0) − f (−1)∣ + ∣f (1) − f (0)∣ = 2. On the other hand, let P = {x0 < ⋯ < xn } be any partition of [−1, 1]. If one extra point w is added to the partition P to make a new partition Q, then V (f ; Q) ≥ V (f ; P ). Indeed, if xk−1 < w < xk , then the Triangle Inequality gives ∣f (xk ) − f (xk−1 )∣ ≤ ∣f (xk ) − f (w)∣ + ∣f (w) − f (xk−1 )∣. This shows that V (f ; P ) ≤ V (f ; Q). Let Q be the partition obtained from P by addition the point 0, if it is not already in P . Since f is decreasing on [−1, 0] and increasing on [0, 1], the telescoping sum computation in Example 4.13 shows that V (f ; P ) ≤ V (f ; Q) = (f (−1) − f (0)) + (f (1) − f (0)) = 2. Since this is true for any partition P of [−1, 1], we conclude that V (f ; [−1, 1]) ≤ 2. Together with the previous paragraph, we have V (f, [−1, 1]) = 2, so f is of bounded variation but not monotone. 4.7.2

Variations on Different Intervals

Theorem 4.14. Suppose f is a function of bounded variation on [a, b], and [c, d] ⊆ [a, b]. Then f is of bounded variation on [c, d], and V (f ; [c, d]) ≤ V (f ; [a, b]). Proof.

Suppose that P = {x0 < ⋯ < xn } is a partition of [c, d]. Then Q = {a ≤ x0 < ⋯ < xn ≤ b}

is a partition of [a, b], where a is inserted into P as the first point in Q if a < c, and b is inserted as the last point in Q if d < b. We have V (f ; P ) ≤ V (f ; Q) ≤ V (f ; [a, b]). Since this is true for any partition P of [c, d], we conclude that V (f ; [c, d]) ≤ V (f ; [a, b]) < ∞, as desired.



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The following result is a sort of converse to Theorem 4.14. It will also be used to characterize functions of bounded variation as differences of increasing functions. Theorem 4.15. Suppose a < c < b, and f is of bounded variation on [a, c] and on [c, b]. Then V (f ; [a, b]) = V (f ; [a, c]) + V (f ; [c, b]),

(4.6)

and f is of bounded variation on [a, b]. Ideas of Proof. The main point is that it is possible to add a point to a given partition, as in Example 4.13. Moreover, the desired equality holds for a fixed partition of [a, b] that includes c. Proof. The assertion that f is of bounded variation on [a, b] will follow from the equality (4.6), since V (f ; [a, c]) and V (f ; [c, b]) are assumed to be finite. To prove (4.6), let P = {x0 < ⋯ < xn } be any partition of [a, b]. The following argument is similar to the one used in Example 4.13. Let Q be the partition of [a, b] obtained from P by adding the point c, if it is not already in P . Then Q gives rise to the partitions Q1 = {a = x0 < ⋯ < c} and Q2 = {c < ⋯ < xn = b} of [a, c] and [c, b], respectively. As in Example 4.13, we have V (f ; P ) ≤ V (f ; Q) = V (f ; Q1 ) + V (f ; Q2 ) ≤ V (f ; [a, c]) + V (f ; [c, b]). Since this is true for any partition P of [a, b], we infer that V (f ; [a, b]) ≤ V (f ; [a, c]) + V (f ; [c, b]) < ∞.

(4.7)

It remains to prove the reverse inequality. Pick any partitions P1 = {x0 < ⋯ < xr = c} and P2 = {c = xr < ⋯ < xr+s } of [a, c] and [c, b], respectively. Splicing these partitions together, we obtain a partition P = {x0 < ⋯ < xr < ⋯ < xr+s } of [a, b] with xr = c. Then we have V (f ; P1 ) + V (f ; P2 ) = V (f ; P ) ≤ V (f ; [a, b]). Since this is true for any partition P1 of [a, c], we have V (f ; [a, c]) + V (f ; P2 ) ≤ V (f ; [a, b]). Likewise, this is true for any partition P2 of [c, b], so V (f ; [a, c]) + V (f ; [c, b]) ≤ V (f ; [a, b]). Combining this inequality with (4.7), we conclude that (4.6) is true.



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Characterization

Now we are ready to show that every function of bounded variation can be written as the difference of two increasing functions. Theorem 4.16. Suppose f is a function of bounded variation on [a, b]. Then there exist increasing functions g, h∶ [a, b] → R such that f = g − h. Proof.

Consider the function g∶ [a, b] → R defined as ⎧ ⎪ if x = a, ⎪0 g(x) = ⎨ (4.8) ⎪ ⎪ ⎩V (f ; [a, x]) if a < x ≤ b. This function g is well-defined by Theorem 4.14. First we observe that g is increasing. For a ≤ c < d ≤ b, Theorem 4.15 implies that g(d) = V (f ; [a, d]) = g(c) + V (f ; [c, d]),

so g(d) − g(c) = V (f ; [c, d]) ≥ 0. Thus, g is increasing. We define h as the difference h = g − f . Then f = g − h, so it remains to show that h is increasing. For a ≤ c < d ≤ b, using the trivial partition {c < d} of [c, d], we have g(d) − g(c) = V (f ; [c, d]) ≥ ∣f (d) − f (c)∣ ≥ f (d) − f (c). So we have h(d) = g(d) − f (d) ≥ g(c) − f (c) = h(c), and h is, therefore, increasing.



Theorem 4.16 allows us to carry certain nice properties of monotone functions to functions of bounded variation. For example, since monotone functions are Riemann integrable, so are functions of bounded variation. The following result gives another example. Corollary 4.2. Suppose f is a function of bounded variation on [a, b]. Then the set of points of discontinuity of f is either finite or countable. Proof. Write f as g −h, where g and h are increasing, hence monotone, functions. Each of g and h has at most countably many points of discontinuity (Theorem 4.12). If f is discontinuous at a point c, then at least one of g and h is discontinuous at c. So the set of points of discontinuity of f is a subset of the union of two finite or countable sets, which is at most countable.  The converse of Theorem 4.16 is also true. In other words, if g, h∶ [a, b] → R are increasing functions, then their difference f = g − h is of bounded variation on [a, b] (Exercise (4) below). Thus, functions of bounded variation are exactly differences of increasing functions. Some nice properties of the class of functions of bounded variation are explored in the exercises.

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Exercises

(1) Let f ∶ [a, b] → R be of bounded variation on [a, b]. Prove that ∣f (x)∣ ≤ ∣f (a)∣ + V (f ; [a, b]) for all x in [a, b]. In particular, f is bounded on [a, b]. (2) Finish the proof of Theorem 4.13 by proving the decreasing case. (3) Let f, g∶ [a, b] → R be of bounded variation on [a, b]. (a) (b) (c) (d)

For a real number c, prove that cf is of bounded variation on [a, b]. Prove that f + g is of bounded variation on [a, b]. Prove that f − g is of bounded variation on [a, b]. Prove that the product f g is of bounded variation on [a, b].

These properties imply that functions of bounded variation on [a, b] form an algebra of functions. In general, an algebra of functions is a set of functions that is closed under scalar multiplication, addition, and product. (4) If g, h∶ [a, b] → R are increasing functions, prove that their difference f = g − h is of bounded variation on [a, b]. (5) Let f ∶ [a, b] → R be of bounded variation on [a, b]. Suppose that there exists a real number r > 0 such that f (x) ≥ r for all x ∈ [a, b]. Prove that f1 is of bounded variation on [a, b]. (6) In each case, compute the variation V (f, [a, b]). (a) f (x) = sin(x) and [a, b] = [0, 2π]. (b) f (x) = cos2 (x) and [a, b] = [0, π]. (c) f (x) = x2 − x and [a, b] = [−2, 2]. (7) Consider the function f defined on [0, π1 ] as ⎧ 1 ⎪ ⎪sin ( x ) if x =/ 0, f (x) = ⎨ ⎪ if x = 0. ⎪ ⎩0 (a) Prove that f is bounded on [0, π1 ]. (b) Prove that f is not of bounded variation on [0, π1 ]. So a bounded function is not necessarily of bounded variation. (c) If 0 < a < π1 , prove that f (x) = sin( x1 ) is of bounded variation on [a, π1 ]. (8) Let f ∶ [a, b] → R be of bounded variation on [a, b]. (a) Prove that limx→c− f exists when a < c ≤ b. (b) Prove that limx→c+ f exists when a ≤ c < b. (9) Let f ∶ [a, b] → R be of bounded variation on [a, b]. Suppose that f is continuous on (a, b). Prove that f is uniformly continuous on (a, b). (10) Let f ∶ [a, b] → R be a function. Suppose that P = {x0 < ⋯ < xn } is a partition of [a, b] such that f is monotone on each subinterval [xk−1 , xk ]. (a) Compute V (f ; [a, b]) in terms of f (xk ) for k = 0, 1, . . . , n. (b) Prove that f is of bounded variation on [a, b].

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(11) Prove that the characteristic function χQ (4.2) is not of bounded variation on [a, b]. (12) Let f ∶ [a, b] → R be a function, and let P be a partition of [a, b]. Suppose that Q is another partition of [a, b] such that all the points in P are also in Q. Prove that V (f ; P ) ≤ V (f ; Q). (13) Let f ∶ [a, b] → R be of bounded variation on [a, b], and let g be defined as in (4.8). (a) Prove that g(x) ≥ 0 for all x ∈ [a, b]. (b) Prove that V (f ; [a, b]) = V (g; [a, b]). (c) If g is continuous at c ∈ (a, b), prove that f is also continuous at c. (14) Let f ∶ [a, b] → R be of bounded variation on [a, b]. Prove that there exist strictly increasing functions f1 , f2 ∶ [a, b] → R such that f = f1 − f2 . (15) Let r be a real number with 0 < r < 1. Consider the function g∶ [0, 1] → R defined as ⎧ ⎪ −rn ⎪ ⎪ ⎪ ⎪ n g(x) = ⎨r ⎪ ⎪ ⎪ ⎪ ⎪ ⎩0

if x = if x =

1 n 1 n

with n odd, with n even,

otherwise.

Prove that g is of bounded variation. (16) Consider the function h∶ [0, 1] → R defined as ⎧ 1 ⎪ ⎪1 if x = n with n even, h(x) = ⎨ ⎪ ⎪ ⎩0 otherwise. Prove that h is not of bounded variation. (17) Show by an example that the composition f ○ g of two functions f and g of bounded variation is not necessarily of bounded variation.

4.8

Additional Exercises

(1) Let f ∶ [a, ∞) → R be a function, and let L be a real number. We write limx→∞ f = L if for every sequence an → ∞ with each an > a, we have f (an ) → L. In this case, we say that the limit of f at ∞ is L. Prove that limx→∞ f = L if and only if for every  > 0, there exists M > 0 with M > a such that x > M implies ∣f (x) − L∣ < . (2) Using the previous exercise as a guide: (a) Write down a reasonable definition of limx→−∞ f = L in terms of sequences. (b) Prove an -M characterization of limx→−∞ f = L. (3) Let f ∶ [a, ∞) → R be a continuous function such that limx→∞ f exists. Prove that f is uniformly continuous on [a, ∞).

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(4) Let f ∶ [a, ∞) → R be a function. We write limx→∞ f = ∞ if for every sequence an → ∞ with each an > a, we have f (an ) → ∞. Prove that limx→∞ f = ∞ if and only if for every α > 0, there exists M > 0 with M > a such that x > M implies f (x) > α. (5) Using the previous exercise as a guide: (a) Write down a reasonable definition of limx→∞ f = −∞ in terms of sequences. Then prove an α-M characterization of limx→∞ f = −∞. (b) Repeat (a) for limx→−∞ f = ∞. (c) Repeat (a) for limx→−∞ f = −∞. (6) Let f ∶ [a, b] → R be a monotone function. Prove that every subinterval [c, d] ⊆ [a, b] contains uncountably many points at which f is continuous. (7) Suppose that a ∈ A ⊆ R. We say that a is an isolated point of A if there exists an open interval J such that J ∩ A = {a}. (a) Prove that a is an isolated point of A if and only if it is not a limit point of A. (b) Let f ∶ A → R be a function. Prove that f is continuous at every isolated point of A. (8) Let f be a polynomial with odd degree. Prove that there exists a ∈ R such that f (a) = 0. (9) Let f, g∶ A → R be two functions. Define h∶ A → R as ⎧ ⎪ ⎪f (x) if f (x) ≥ g(x), h(x) = max{f (x), g(x)} = ⎨ ⎪ ⎪ ⎩g(x) if f (x) < g(x). (a) If f and g are both continuous at a ∈ A, prove that h is continuous at a. (b) Prove that F (x) = max{f (x), 0} is continuous at any point at which f is continuous. (10) Let f ∶ [0, 2] → R be a continuous function such that f (0) = f (2). Prove that there exists a ∈ [0, 1] such that f (a) = f (a + 1). (11) Let f ∶ [a, b] → R be a continuous function. Suppose that the set f ([a, b]) is either finite or countable. Prove that f is a constant function, i.e., f (x) = c for all x ∈ [a, b] for some fixed real number c. (12) Let f ∶ A → R be a uniformly continuous function on a bounded set A. (a) Prove that f is bounded on A. (b) Show by examples that f may not be bounded on A if it is merely assumed to be continuous or if A is not bounded. (13) A function f ∶ R → R is said to be p-periodic if there exists p > 0 such that f (x + p) = f (x) for all x ∈ R. For example, sin(x) is 2π-periodic with p = 2π. Prove that a continuous periodic function is uniformly continuous and bounded on R. (14) Let f ∶ [a, b] → R be a continuous function such that f (a) = f (b). Prove that there exists a periodic uniformly continuous function g∶ R → R such that g(x) =

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f (x) for x ∈ [a, b]. Such a function g is called a periodic extension of f to R. (15) A function f ∶ A → R is called a Lipschitz function if there exists a real number M > 0 such that ∣f (a) − f (b)∣ ≤ M ∣a − b∣ for all a, b ∈ A. (a) Prove that a Lipschitz function is uniformly continuous. √ (b) Prove that f (x) = x∶ [c, ∞) → R for any c > 0 is a Lipschitz function. (c) Give an example of a uniformly continuous function that is not a Lipschitz function. (16) In each case, determine whether the function f ∶ A → R is a Lipschitz function. (a) (b) (c) (d)

f (x) = x2 and A = [a, b]. f (x) = x2 and A = [0, ∞). f (x) = x1 and A = [1, b] for some b > 1. 1 f (x) = 1+x 2 and A = R.

(17) Let f, g∶ A → R be Lipschitz functions. (a) Prove that f + g∶ A → R is a Lipschitz function. (b) Suppose, in addition, that f and g are bounded on A. Prove that the product f g∶ A → R is a Lipschitz function. (18) Consider the function f ∶ (0, ∞) → R defined as ⎧ 1 ⎪ ⎪ f (x) = ⎨ q ⎪ ⎪ ⎩0

if x ∈ Q with x =

p q

in lowest terms,

if x ∈/ Q.

(a) Prove that f is discontinuous at every rational x ∈ (0, ∞). (b) Prove that f is continuous at every point x ∈/ Q with x ∈ (0, ∞). This is called the Thomae function. (19) Consider the function f ∶ [0, 1] → [0, 1] defined as ⎧ ⎪ if x ∈ Q, ⎪x f (x) = ⎨ ⎪ 1 − x if x ∈/ Q. ⎪ ⎩ (a) Prove that f is a bijection. (b) Prove that f is not monotone on any subinterval of [0, 1]. (c) Prove that f is continuous at 21 and discontinuous on [0, 12 ) ∪ ( 21 , 1]. This exercise illustrates that a function defined on a closed bounded interval may satisfy the conclusion of the Intermediate Value Theorem without being continuous on any subinterval. (20) Let f ∶ A → R be a function, and suppose that a ∈ A. We say that f is left continuous at a if for every sequence an → a with each an ∈ A and an < a, the sequence {f (an )} converges to f (a).

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(a) Prove that f is left continuous at a if and only if for every  > 0, there exists δ > 0 such that 0 ≤ a − x < δ with x ∈ A implies ∣f (x) − f (a)∣ < . (b) Give a reasonable definition of right continuous at a. (c) Prove an -δ characterization of right continuity. (d) Prove that f is continuous at a if and only if f is left continuous at a and right continuous at a. (21) Give an example in which a function f is left continuous at a point a but is not right continuous at a. (22) Consider the function f ∶ [1, 3] → R defined as 1 ⎧ ⎪ 4x + 1 2 ⎪ ⎪ ⎪( ) f (x) = ⎨ 14x − 3 ⎪ ⎪ ⎪ ⎪ ⎩3x − 2

if 1 ≤ x < 2, if 2 ≤ x ≤ 3.

Determine if f is left continuous at 2, right continuous at 2, or neither. (23) Let {xn } be a strictly increasing convergent sequence, and let ∑ an be a convergent series with each an > 0. Define a function f ∶ R → R as ∞

f (x) = ∑ an J(x − xn ), n=1

where J(x) is the step function from Exercise (10) on p. 104. (a) (b) (c) (d)

Prove Prove Prove Prove

that that that that

f is a well-defined increasing function. f (x) = 0 for x < x1 and f (x) = ∑ an for x ≥ lim xn . f is continuous on R ∖ {xn ∶ n ∈ Z+ }. f is discontinuous at each xn .

(24) A subset A ⊆ R is open if for each a ∈ A, there exists δ > 0 such that the open interval (a − δ, a + δ) is a subset of A. A subset A ⊆ R is closed if R ∖ A is open. These concepts were introduced in Exercises (14) – (18) on p. 26. Prove that A ⊆ R is closed if and only if A contains all of its limit points. (25) A subset S ⊆ A is said to be open in A if there exists an open set O such that S = A ∩ O. Let f ∶ A → R be a function. (a) Prove that f is continuous on A if and only if for every open set U ⊆ R, the set f −1 (U ) is open in A. (b) If f ∶ (a, b) → R is continuous and strictly monotone, prove that f (U ) is open for every open U in (a, b). (c) Show by an example that the conclusion of the previous part is not necessarily true if f is not required to be strictly monotone. (26) In this exercise, you will show that a subset A ⊆ R is open if and only if it is the disjoint union of at most countably many open intervals. (a) Prove that a disjoint union of finitely or countably many open intervals is open.

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(b) Let A ⊆ R be open and non-empty. For each a ∈ A, consider the set Ia = ⋃ J, where J ⊆ A is an open interval containing a. Prove that Ia is an open interval containing a with Ia ⊆ A. (c) If A is open and a, b ∈ A, prove that Ia ∩ Ib =/ ∅ implies Ia = Ib . (d) Using the previous two parts, prove that if A is open, then A is the disjoint union of at most countably many open intervals. (27) Let f ∶ A → R be a continuous and bounded function on a closed set A. In this exercise, you will prove that there exists a continuous function g∶ R → R such that g(x) = f (x) for all x ∈ A. Since R ∖ A is open, by the previous exercise, it is the disjoint union of at most countably many open intervals Ik = (ak , bk ) (or (−∞, bk ) or (ak , ∞)) with end points in A. Define g∶ R → R as ⎧ f (x) ⎪ ⎪ ⎪ ⎪ ⎪ f (bk ) − f (ak ) ⎪ ⎪ ⎪ ⎪f (ak ) + (x − ak ) bk − ak g(x) = ⎨ ⎪ ⎪ ⎪ f (b ) k ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ f (a k) ⎩

if x ∈ A, if x ∈ Ik = (ak , bk ), if x ∈ Ik = (−∞, bk ), if x ∈ Ik = (ak , ∞).

Prove that g is continuous on R. This is a special case of the Tietze Extension Theorem. The function g is called a continuous extension of f to R.

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Chapter 5

Differentiation

The derivative of a function measures its rate of change. In terms of its graph, the derivative of a nice function can be interpreted as the slope of the graph. It is a very important concept in analysis and in the sciences. Basic rules regarding the computation of derivatives are discussed in section 5.1. In section 5.2 we discuss the Mean Value Theorem, which roughly says that the slope of any secant line of the graph of a function can be achieved as the derivative at some point. Several important results are consequences of the Mean Value Theorem, including Taylor’s Theorem, which is discussed in section 5.3, and the Fundamental Theorems of Calculus.

5.1

The Derivative

We assume that the reader is familiar with the interpretations of the derivative as a rate of change or, geometrically, as the slope of the tangent line to the graph. Thus, we will concentrate on proving properties and consequences of the derivative in details. We begin with its definition. Recall that an open interval is an interval of the form (a, b), (−∞, b), (a, ∞), or (−∞, ∞). 5.1.1

Definition of Derivative

Definition 5.1. Let f ∶ I → R be defined on an open interval I, and let c ∈ I. The derivative of f at c is defined as the limit f ′ (c) = lim

x→c

f (x) − f (c) , x−c

if it exists. If this is the case, we say that f is differentiable at c. If this limit does not exist, then we say that f is not differentiable at c. If f is differentiable at every point in I, then we say that f is differentiable on I and write f ′ for the function whose value at x ∈ I is f ′ (x). Similarly, write f (n) (c) = (f (n−1) )′ (c), 115

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if the derivative exists, and call it the nth derivative of f at c. The second and the third derivatives are also written as f ′′ and f ′′′ , respectively, if they exist. The 0th derivative f (0) is defined as f itself. We say that f is continuously differentiable if f ′ exists and is continuous. If f (n) exists for all n, then f is called infinitely differentiable. A function f is not differentiable at c if and only if there exists a sequence {xn } ∈ I ∖ {c} converging to c such that the sequence f (xn ) − f (c) } xn − c does not converge. From now on, when we say f is differentiable at c, we automatically assume that f is defined on some open interval containing c. The alternative df is also used for f ′ (x). notation dx The limit of a function was defined in Definition 4.2 in terms of sequences. An -δ characterization of limit was given in Theorem 4.3. As an exercise, the reader should write down in details these two equivalent formulations of the derivative. {

5.1.2

Differentiability and Continuity

The first important property of differentiability is that it implies continuity. Theorem 5.1. Let f ∶ I → R be defined on an open interval I, and let c ∈ I. If f is differentiable at c, then f is continuous at c. Proof. Let {xn } be a sequence in I ∖{c} with lim xn → c. Since f is differentiable at c, we have f (xn ) − f (c) lim = f ′ (c). n→∞ xn − c Therefore, we have f (xn ) − f (c) f (xn ) − f (c) = ( ) (xn − c) → f ′ (c) ⋅ 0 = 0, xn − c i.e., lim f (xn ) = f (c). So f is continuous at c.



The converse of the above theorem is not true, as the following example illustrates. Example 5.1. Consider the function f (x) = ∣x∣ defined on R. This function is continuous on R. However, it is not differentiable at 0. Indeed, let xn = n1 if n > 1 is even and xn = − n1 if n ≥ 1 is odd. Then xn → 0, and f (xn ) = n1 for all n. So we have ⎧ ⎪1 if n is even, f (xn ) − f (0) ⎪ =⎨ ⎪−1 if n is odd. xn − 0 ⎪ ⎩ (0) does not exist, and f is not differentiable at 0. This shows that lim f (xxnn)−f −0

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What makes the function f (x) = ∣x∣ not differentiable at 0 is the corner on the graph of f at x = 0. Using a combination of functions with more and more corners on their graphs, it is, in fact, possible to construct a function g that is continuous everywhere but is nowhere differentiable. We will present such an example in a later chapter when we have the machinery of series of functions at our disposal. Be careful that Theorem 5.1 does not say that f ′ is continuous at c. It also does not say that f is uniformly continuous, even if f is differentiable on I. These issues are explored further in the exercises. 5.1.3

Arithmetics of Derivatives

As in the case of sequences, it is convenient to know how derivative behaves with respect to arithmetic operations. Theorem 5.2. Let f, g∶ I → R be defined on an open interval I, and let c ∈ I. Suppose that both f and g are differentiable at c. Then we have: (1) (2) (3) (4)

(af )′ (c) = af ′ (c) for every a ∈ R. (f + g)′ (c) = f ′ (c) + g ′ (c). Product Rule: (f g)′ (c) = f (c)g ′ (c) + f ′ (c)g(c). Quotient Rule: If g(c) =/ 0, then f ′ (c)g(c) − f (c)g ′ (c) f ′ . ( ) (c) = g g(c)2

In the first case, when we write (af )′ (c) = af ′ (c), we mean that the function af is differentiable at c and that its derivative at c is af ′ (c). Similar remarks apply to the other cases. Proof. We will prove the last case, which is the most difficult of the four cases, and leave the first three cases to the reader as an exercise. So suppose that g(c) =/ 0. Since g is also continuous at c (Theorem 5.1), there exists an open interval J ⊆ I containing c such that g(x) =/ 0 for each x ∈ J. Pick any sequence {xn } in I ∖ {c} with xn → c. Then there exists a positive integer N such that n ≥ N implies xn ∈ J, g(xn ) =/ 0, and ( fg ) (xn ) − ( fg ) (c) xn − c

f (xn )g(c) − f (c)g(xn ) (xn − c)g(xn )g(c) (f (xn ) − f (c)) g(c) − f (c) (g(xn ) − g(c)) = (xn − c)g(xn )g(c)

=

f (xn ) − f (c) 1 f (c) g(xn ) − g(c) ) −( )( ) xn − c g(xn ) g(xn )g(c) xn − c f ′ (c) f (c)g ′ (c) − . → g(c) g(c)2

=(

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In the last step, we used the differentiability of f and g at c and the continuity of g at c. We obtain the desired expression for ( fg )′ (c) when we write the above difference as one fraction.  Example 5.2. Consider the function f (x) = xn defined on R, where n is a positive integer. We show by induction that f is differentiable on R and that f ′ (c) = ncn−1 for any real number c. If n = 1, then f (x) − f (c) = x − c, so f ′ (c) = limx→c 1 = 1c0 . Suppose that (xn )′ (c) = ncn−1 for some n ≥ 1. Then (xn+1 )′ (c) = (xn ⋅ x)′ (c) = (xn )(c) ⋅ 1 + (xn )′ (c) ⋅ (c) = cn + ncn−1 ⋅ c = (n + 1)cn . Thus, by induction we have shown that f ′ (x) = nxn−1 for f (x) = xn , where n is any positive integer and x ∈ R. 5.1.4

Chain Rule

The next derivative rule is about √ the composition of two functions. To motivate it, consider the function h(x) = 1 − x2 . Computing its derivative using the definition is quite inconvenient. However, note that h is the composition g○f , where the inside √ function is f (x) = 1 − x2 and the outside function is x. It is not difficult at all to compute the derivatives of f and g using the definition. The reader should give it a try. This leads naturally to the question: Is there a way to express the derivative of a composition in terms of the individual functions and their derivatives? The answer is yes, and that is the content of the following result. Theorem 5.3 (Chain Rule). If f is differentiable at c and g is differentiable at f (c), then the composition g ○ f is differentiable at c and (g ○ f )′ (c) = g ′ (f (c)) ⋅ f ′ (c). Ideas of Proof. The plan is to write down two suitable functions whose limits as x → c are g ′ (f (c)) and f ′ (c), respectively. Proof. Suppose I is an open interval in Dom(f ) containing c, and define the function ϕ by ⎧ g(y) − g(f (c)) ⎪ ⎪ ⎪ ⎪ y − f (c) ϕ(y) = ⎨ ⎪ ′ ⎪ ⎪ g (f (c)) ⎪ ⎩

if y ∈ Dom(g) and y =/ f (c), if y = f (c).

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The differentiability of g at f (c) means g(y) − g(f (c)) = lim ϕ(y), y→f (c) y→f (c) y − f (c)

ϕ(f (c)) = g ′ (f (c)) = lim

which in turn means ϕ is continuous at f (c). From the definition of ϕ, we have g(f (x)) − g(f (c)) f (x) − f (c) = ϕ(f (x)) ⋅ (5.1) x−c x−c for x ∈ Dom(g ○ f ) with x =/ c. Note that (g ○ f )′ (c) is the limit as x → c of the left-hand side of (5.1). Moreover, the limit as x → c of the second factor on the right-hand side of (5.1) is f ′ (c). Since f is also continuous at c, we have limx→c f (x) = f (c). The continuity of ϕ at f (c) now implies lim ϕ(f (x)) = ϕ(f (c)) = g ′ (f (c)),

x→c

which proves the theorem.



Example 5.3. Using the limit definition, one finds that if f (x) = 1 − x2 and g(x) = √ 1 for x > 0. Therefore, chain rule says that the x, then f ′ (x) = −2x and g ′ (x) = 2√ x √ 2 derivative of h(x) = 1 − x = (g ○ f )(x) is −2x h′ (x) = g ′ (f (x)) ⋅ f ′ (x) = √ 2 1 − x2 for −1 < x < 1. 5.1.5

Derivatives of Inverse Functions

In Theorem 4.11 we saw that a strictly monotone continuous function defined on an interval has a strictly monotone continuous inverse function. The following result says that if the function is differentiable with non-zero derivative, then its inverse function is also differentiable. Theorem 5.4 (Inverse Function Theorem). Let f be a strictly monotone continuous function defined on an open interval I, c be a point in I, and f be differentiable at c with f ′ (c) =/ 0. Then its inverse function g = f −1 is differentiable at d = f (c) and 1 g ′ (d) = ′ . f (c) Ideas of Proof. The main point is that f ′1(c) can be computed as the limit of the reciprocal of the difference quotient that appears in the definition of f ′ (c). Proof. The domain J = f (I) of g is an open interval (Exercise (4) on page 96). Given  > 0 we need to show that there exists δ > 0 such that 0 < ∣y − d∣ < δ implies ∣

1 g(y) − c 1 g(y) − g(d) − ′ ∣=∣ − ∣ < . y−d f (c) f (g(y)) − f (c) f ′ (c)

(5.2)

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The assumptions that f is strictly monotone and that f ′ (c) =/ 0 imply the equality 1 f ′ (c)

= lim

x→c

x−c f (x) − f (c)

by Exercise (2d) on page 89. Therefore, with the given , there exists γ > 0 such that 0 < ∣x − c∣ < γ implies ∣

1 x−c − ∣ < . f (x) − f (c) f ′ (c)

(5.3)

Since g is continuous on J (Theorem 4.11), there exists δ > 0 such that 0 < ∣y − d∣ < δ implies 0 < ∣g(y) − g(d)∣ = ∣g(y) − c∣ < γ. The previous inequality implies (5.3) with x = g(y), which is the desired inequality (5.2).  The derivative formula for the inverse function can be derived using chain rule. In fact, since x = g(f (x)), differentiating both sides we obtain 1 = g ′ (f (x)) ⋅ f ′ (x). Writing y = f (x), we then obtain g ′ (y) =

1 . f ′ (x)

However, the reader is cautioned that this line of reasoning does not prove that the inverse function is differentiable, which is what the above theorem proved. In fact, in using chain rule, we are already assuming that g is differentiable. Example 5.4. For a positive integer n, the function f (x) = xn is strictly increasing and differentiable on R if n is odd and on (0, ∞) if n is even. Its derivative is 1 f ′ (x) = nxn−1 . Its inverse function is the nth root function g(x) = x n , which is only defined for x > 0 if n is even. Writing 0 =/ y = f (x) = xn we have g ′ (y) = nx1n−1 , which yields g ′ (x) =

1 n−1 n

=

1 1−n 1 1 x n = x n −1 . n n

nx Therefore, the derivative formula for xn , where n is a positive integer, still holds if n is replaced by n1 . 5.1.6

Exercises

(1) Use the definition of derivative to find the derivatives, when they exist, of the following functions. √ (a) 2x + 5 1 (b) √x+4 (c) 2x3 + 10x + 5 (d) x21+1

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x x2 +1

(2) Write down the sequential and the -δ characterizations of the derivative. Write down what it means for f to not be differentiable at c using -δ. (3) Prove that a constant function is differentiable on R, and compute its derivative 0. (4) Prove the first three parts of Theorem 5.2. (5) Prove that a polynomial is differentiable on R. (6) Suppose r = pq is a rational number in lowest terms and g(x) = xr . Prove that g is differentiable on R ∖ {0} if q is odd and on (0, ∞) if q is even. Then show that g ′ (x) = rxr−1 . (7) Suppose f and g are both differentiable at c and that f (x) ≤ g(x) for all x in some open interval containing c. Does it follow that f ′ (c) ≤ g ′ (c)? (8) Prove the generalized product rule n n (f g)(n) (c) = ∑ ( )f (k) (c)g (n−k) (c), k=0 k assuming all the derivatives exist, where (nk) is the binomial coefficient in (1.4). (9) Define the function ⎧ 1 ⎪ ⎪x sin ( x ) if x =/ 0, f (x) = ⎨ ⎪ if x = 0. ⎪ ⎩0 Prove that f is not differentiable at 0. (10) Define the function ⎧ 1 ⎪ ⎪x2 sin ( x ) if x =/ 0, f (x) = ⎨ ⎪ if x = 0. ⎪ ⎩0 Prove that f is differentiable at 0, and compute f ′ (0). You may assume that the function sin(x) is differentiable on R and that its derivative is cos(x). Show that f ′ is not continuous at 0. (11) Prove that f (x) = ∣x∣ is differentiable on (−∞, 0) and (0, ∞), and compute its derivative. (12) If f is differentiable at c, prove that f (c + h) − f (c) f (c + h) − f (c − h) f ′ (c) = lim = lim . h→0 h→0 h 2h Draw pictures to visualize these limits in terms of secant lines. (13) In the previous exercise, if the last limit exists, does it follow that f is differentiable at c? (14) Let f ∶ R → R be the function f (x) = x2 if x is rational and f (x) = 0 if x is irrational. Prove that f is differentiable at 0, and compute f ′ (0). (15) Give an example of a differentiable function f on an open interval I that is not uniformly continuous. (16) Suppose f is differentiable on an open interval I and that f ′ is bounded on I. Prove that f is uniformly continuous on I.

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Mean Value Theorem

The title of this section refers to an important theorem that says that the slope of the secant line of a function on a closed interval is, in fact, the slope of some tangent line. 5.2.1

Two Preliminary Theorems

In order to prove the Mean Value Theorem, we need two preliminary observations. The following observation says that if a function achieves an extremum at a point where it is differentiable, then its derivative must be zero there. Intuitively, this is true because, if the derivative is not zero, then the function is either increasing or decreasing there. But then that point cannot be an extremum. The proof is a formal version of this line of reasoning. Theorem 5.5 (Interior Extremum Theorem). Suppose f is defined on an open interval I, c is a point in I where f is differentiable, and f achieves its maximum or minimum on I at c. Then f ′ (c) = 0. Ideas of Proof. If f ′ (c) =/ 0, then there are points close to c where f is larger or smaller than f (c), which would contradict the maximum or minimum assumption.

Proof. Say f achieves its maximum at c. The minimum case is left as an exercise. Suppose to the contrary that f ′ (c) =/ 0, say, f ′ (c) > 0. Then there exists δ > 0 such that f (x) − f (c) > 0. 0 < ∣x − c∣ < δ with x ∈ I implies x−c Choosing any x > c in I, the above inequality then implies f (x) > f (c), which contradicts the maximum assumption. If f ′ (c) < 0, then likewise there exists δ > 0 such that f (x) − f (c) 0 < ∣x − c∣ < δ with x ∈ I implies < 0. x−c Choosing any x < c in I, the above inequality implies f (x) > f (c), which is again a contradiction.



The following observation is a special case of the Mean Value Theorem. It says roughly that if a function has equal values at two distinct points, then its derivative must be zero somewhere between those two points. Intuitively, this is easy to see if one tries to draw a graph of such a function.

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Theorem 5.6 (Rolle’s Theorem). Suppose f is continuous on a closed interval [a, b], is differentiable on (a, b), and f (a) = f (b). Then there exists a point c ∈ (a, b)

such that

f ′ (c) = 0.

Proof. By the Interior Extremum Theorem, if f achieves either a maximum or a minimum on (a, b) at a point c, then f ′ (c) = 0. By the Extreme Value Theorem 4.6, f achieves both a maximum and a minimum on [a, b]. If either one of them is achieved on (a, b), then we are done by the Interior Extremum Theorem. Otherwise, f achieves its maximum and minimum at a or b. But then the assumption f (a) = f (b) implies that f is constant on [a, b]. Therefore, f ′ (c) = 0 for every point c in (a, b).  5.2.2

Main Result

We are now ready for the main result of this section. Theorem 5.7 (Mean Value Theorem). Suppose f is continuous on a closed interval [a, b] and differentiable on (a, b). Then there exists a point c in (a, b) such that f ′ (c) =

f (b) − f (a) . b−a

Ideas of Proof. The plan is to use Rolle’s Theorem by subtracting from f the secant line connecting the points at x = a and x = b, whose equation is y = f (b) +

Proof.

f (b) − f (a) (x − b). b−a

Define the function g(x) = f (x) − f (b) −

f (b) − f (a) (x − b), b−a

which is continuous on [a, b] and differentiable on (a, b). Since g(a) = g(b) = 0, Rolle’s Theorem implies that there exists a point c in (a, b) such that 0 = g ′ (c) = f ′ (c) − as desired.

f (b) − f (a) , b−a 

The reader should draw a typical graph and try to interpret the Mean Value Theorem using the graph. In any case, the quotient on the right-hand side is the slope of a secant line, and f ′ (c) is the slope of a tangent line somewhere in between.

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Note that Rolle’s Theorem is a special case of the Mean Value Theorem because the quotient in the latter is 0 when f (a) = f (b). The Mean Value Theorem, and its special case Rolle’s Theorem, are ultimately about showing that certain quantity is in the range of the derivative. One way of using them is to establish certain inequalities that are otherwise hard to prove, as the following example illustrates. Example 5.5. Here we show that ex ≤ ex for all x in R. If f (x) = ex − ex, then we must show f (x) ≥ 0. Note that the only root of f (x) = 0 is x = 1. Indeed, if there is another root r =/ 1, then Rolle’s Theorem says f ′ (d) = 0 for some point d in between r and 1. But the only root of f ′ (x) = ex − e = 0 is x = 1. Moreover, f ′ (x) < 0 if x < 1 and f ′ (x) > 0 if x > 1, so f is strictly decreasing on (−∞, 1) and strictly increasing on (1, ∞). Since x = 1 is the only zero of f , we conclude that f (x) ≥ 0 for all x. 5.2.3

Consequences

We now discuss some consequences of the Mean Value Theorem. The following observation says that a function whose derivative is identically 0 is a constant function. Intuitively, having derivative 0 implies that the function can neither increase nor decrease, which means it is constant. Corollary 5.1. Suppose f is differentiable on (a, b) with f ′ = 0 on (a, b). Then f is a constant function. Proof. If f is not a constant function, then there exist c < d in (a, b) such that f (c) =/ f (d). Applying the Mean Value Theorem to f on [c, d], we obtain a point e in (c, d) such that f ′ (e) =

f (d) − f (c) =/ 0, d−c

which contradicts the assumption f ′ = 0.



The next observation says that if two functions have the same derivative on some interval, then they can differ by at most a constant. Intuitively, these two functions have to increase or decrease at exactly the same rate everywhere on this interval. So they differ by the same amount at all the points on this interval. Corollary 5.2. Suppose f and g are both differentiable on (a, b) with f ′ = g ′ on (a, b). Then there exists a constant C such that f = g + C on (a, b).

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Proof. The difference h = f − g is differentiable on (a, b) and satisfies h′ = 0, so the previous corollary says h is a constant function. In other words, f = g + C for some constant C.  The previous result is important in integration. It says that if a function f has an anti-derivative F on an interval, then {F + C ∶ C constant} is the entire family of anti-derivatives of f . Therefore, to find all the anti-derivatives of a function, it suffices to find just one of them and form the above family. More consequences of the Mean Value Theorem are explored in the exercises and the next section. 5.2.4

Exercises

(1) Finish the proof of Theorem 5.5 by proving the case when f achieves its minimum at c. (2) Give a proof of Bernoulli’s Inequality (Theorem 1.5) using the Mean Value Theorem. (3) Prove the inequality ∣ cos x − cos y∣ ≤ ∣x − y∣ for all real numbers x and y. (4) Prove the inequality ∣ sin x − sin y∣ ≤ ∣x − y∣ for all real numbers x and y. (5) Suppose f is differentiable on an open interval I. Prove the following statements. (a) (b) (c) (d)

If If If If

f ′ (x) > 0 f ′ (x) ≥ 0 f ′ (x) < 0 f ′ (x) ≤ 0

for for for for

all all all all

x x x x

in in in in

I, I, I, I,

then then then then

f f f f

is is is is

strictly increasing. increasing. strictly decreasing. decreasing.

(6) Suppose f and g are both differentiable, and f (a) = g(a). (a) If f ′ (x) ≤ g ′ (x) for all x ≥ a, prove that f (x) ≤ g(x) for all x ≥ a. (b) If f ′ (x) < g ′ (x) for all x > a, prove that f (x) < g(x) for all x > a. (7) (8) (9) (10) (11)

Prove that ln(1 + x) < x for all x > 0. Prove that ∣ sin x∣ ≤ x for all x ≥ 0. Prove that tan x > x if 0 < x < π2 . Prove that ex > 1 + x for all x > 0. Suppose f is differentiable on an open interval I with f ′ bounded. Prove that there exists a real number M > 0 such that ∣f (x) − f (y)∣ ≤ M ∣x − y∣

for all x, y in I. (12) Suppose f is continuous on [a, b], differentiable on (a, b), and f (c) > f (a) = f (b) for some point c in (a, b). Prove that there exist two points x and y in (a, b) such that f ′ (x) < 0 < f ′ (y). (13) Suppose f is continuous on [0, 1], differentiable on (0, 1), f (0) = 0, and f ′ is increasing on (0, 1). Prove that g(x) = f (x) is increasing on (0, 1). x

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1

1

(14) Suppose 0 < x < y and n ≥ 2 is an integer. Prove that y n − x n < (y − x) n . (15) Suppose x, y > 0 and 0 < r < 1. Prove that xr y 1−r ≤ rx + (1 − r)y. (16) Suppose f is continuously differentiable on an open interval I and a is a point in I such that f (a) = 0 and f ′ (a) =/ 0. Prove that there exists an open interval J ⊆ I containing a on which both f and f ′ are non-zero with the exception of f (a) = 0. 5.3

Taylor’s Theorem

In this section we discuss other important results related to the Mean Value Theorem, the main one being Taylor’s Theorem. We then discuss L’Hospital’s Rule and the intermediate value property for derivatives. 5.3.1

Motivation

Let us discuss the motivation behind the statement of Taylor’s Theorem. Polynomials are among the easiest functions to work with. For example, they have very simple derivative formulas, and the product of two polynomials remains a polynomial. Moreover, computing a polynomial involves only the arithmetic operations of addition, subtraction, and multiplication, which can be easily handled by computers. Therefore, it makes sense to approximate other functions using polynomials. Moreover, if we want to understand a function f near a point c, then the polynomials should be in powers of (x − c) instead of x. Suppose it is possible to write f (x) = a0 + a1 (x − c) + a2 (x − c)2 + a3 (x − c)3 + ⋯ in powers of (x − c). Then f (c) = a0 ,

f ′ (c) = a1 ,

f ′′ (c) = 2a2 ,

f ′′′ (c) = 3!a3 ,

and so forth. The general formula for the kth coefficient is f (k) (c) . k! Of course, f does not need to be a polynomial, so the above expression of f should have an error term. In other words, a degree n polynomial approximation of f should take the form ak =

f (x) = Tn (x; c) + Rn (x; c),

(5.4)

where Tn (x; c) = f (c) + f ′ (c)(x − c) + ⋯ + n

f (k) (c) (x − c)k k! k=0

=∑

f (n) (c) (x − c)n n!

(5.5)

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is a degree n polynomial in (x − c), and Rn (x; c) is the nth error term. The polynomial Tn (x; c) is called the degree n Taylor polynomial of f at c. Taylor’s Theorem provides an actual expression for the error term in terms of a higher derivative. 5.3.2

Main Result

We now state and prove Taylor’s Theorem. Theorem 5.8. Suppose n is a positive integer and f ∶ [α, β] → R is a function such that f (k) is continuous on [α, β] for 0 ≤ k ≤ n and that f (n+1) exists on (α, β). Then given two distinct points c and x in [α, β], there exists a point b strictly between them such that f (x) = Tn (x; c) +

f (n+1) (b) (x − c)n+1 . (n + 1)!

Ideas of Proof. We are trying to show that certain quantity is in the range of the higher derivative f (n+1) . Therefore, as in the proof of the Mean Value Theorem, we are going to use Rolle’s Theorem on a suitable function. Proof.

We need to show that, in the equality K (x − c)n+1 = f (x) − Tn (x; c), (n + 1)!

the quantity K must be f (n+1) (b) for some b strictly between c and x. To prove this, consider the function in t F (t) = f (x) − Tn (x; t) −

K (x − t)n+1 (n + 1)!

defined on the closed interval J formed by c and x. The function F (t) is continuous on J, differentiable between c and x, and satisfies F (c) = 0 = F (x) by the definition of K and by Tn (x; x) = f (x). Therefore, Rolle’s Theorem 5.6 yields a point b strictly between c and x such that F ′ (b) = 0. The reader can check with a direct computation that d f (n+1) (t) (Tn (x; t)) = (x − t)n . dt n!

(5.6)

Therefore, we have 0 = F ′ (b) = −

K(n + 1) f (n+1) (b) (x − b)n − (x − b)n (−1), n! (n + 1)!

which simplifies to K = f (n+1) (b), as desired.



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A First Course in Analysis

In terms of the error term in (5.4), Taylor’s Theorem says f (n+1) (b) (x − c)n+1 (5.7) Rn (x; c) = (n + 1)! for some point b between c and x. In other words, the error term is given in terms of the (n + 1)st derivative of f . Another form of the error term is discussed in Theorem 6.10. One typical use of Taylor’s Theorem is to approximate functions using low degree polynomials, as in the next example. Example 5.6. Suppose f (x) = sin x and a = 0. For any x =/ 0, Taylor’s Theorem with n = 3 yields f (4) (c) 4 x3 sin c 4 sin x = T3 (x; 0) + x =x− + x 4! 3! 4! for some c strictly between 0 and x. Since ∣ sin x∣ ≤ 1 for all x, the error term is 4 bounded above by x4! , which is a very small number for x near 0. For example, if x = 0.1, then

5.3.3

(0.1)4 4!

is about 4.2 × 10−6 .

L’Hospital’s Rule

Next we discuss a simple version of L’Hospital’s Rule, which the reader must have encountered in differential calculus. The result says roughly that, if f (a) = 0 = g(a), ′ (a) if it exists. then limx→a fg is equal to fg′ (a) Theorem 5.9. Suppose f and g are both continuously differentiable on an open interval I, a is a point in I such that f (a) = 0 = g(a), and that g ′ (a) =/ 0. Then there is an equality f (x) f ′ (a) lim = ′ . x→a g(x) g (a) Ideas of Proof. Near the point a, the graphs of f and g are approximated by their respective tangent lines. Since f (a) = 0 = g(a), these lines have equations y = f ′ (a)(x − a) and y = g ′ (a)(x − a), respectively, which are also the degree 1 Taylor polynomials of f and g. Therefore, near the point a, the quotient fg is roughly equal to f ′ (a)(x − a) f ′ (a) = . g ′ (a)(x − a) g ′ (a) Proof. There is an open interval J ⊆ I containing a on which g and g ′ are nonzero with the exception of g(a) = 0 (Exercise (16) on page 126). For x =/ a in J, the Mean Value Theorem says f (x)−f (a) f (x) ( x−a ) f ′ (c) = = g(x) ( g(x)−g(a) ) g ′ (d) x−a

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for some c and d strictly between a and x. To finish the proof, just take the limit limx→a , using the assumption that f ′ and g ′ are continuous.  Example 5.7. We have lim

x→1

by L’Hospital’s Rule, since (ln x)′ =

1 x

ln x =1 x−1

and (x − 1)′ = 1.

There is a more general form of L’Hospital’s Rule involving higher derivatives. It will be explored in the exercises. 5.3.4

Intermediate Value Theorem for Derivative

Recall that a continuous function on a closed bounded interval satisfies the Intermediate Value Theorem 4.7. On the other hand, a differentiable function does not need to have a continuous derivative, so the Intermediate Value Theorem does not apply to the derivative in general. Nevertheless, the next observation says that the derivative does satisfy the conclusion of the Intermediate Value Theorem. Its proof uses the Interior Extremum Theorem instead of the Mean Value Theorem. Theorem 5.10. Suppose f is differentiable on (a, b), and r is a number that lies strictly between f ′ (c) and f ′ (d) for some c < d in (a, b). Then there exists a point e ∈ (c, d)

such that

r = f ′ (e).

Ideas of Proof. The plan is similar to the proof of the Mean Value Theorem. We will subtract from f the line through the origin with slope r and apply the Interior Extremum Theorem. Proof. Let us consider the case f ′ (c) < r < f ′ (d). The other case is very similar. Consider the function g(x) = f (x) − rx, which is differentiable on (a, b) with g ′ (x) = f ′ (x) − r. In particular, it suffices to show that there exists a point e in (c, d) with g ′ (e) = 0. Thus, by the Interior Extremum Theorem, it is enough to show that g achieves either a maximum or a minimum on (c, d) at some point e. By the Extreme Value Theorem 4.6, g achieves a minimum on [c, d] at some point e. To apply the Interior Extremum Theorem, we just need to observe that e is neither c nor d. In other words, we need to show that g(x1 ) < g(c)

and g(x2 ) < g(d)

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for some x1 and x2 in (c, d). Note that g ′ (c) < 0 < g ′ (d). So there exists δ1 > 0 such that 0 < ∣x − c∣ < δ1 with x ∈ (a, b)

implies

g(x) − g(c) < 0, x−c

which in turn implies g(x) < g(c)

if x > c.

Therefore, e is not the end point c. Likewise, the inequality 0 < g ′ (d) implies that there exists δ2 > 0 such that 0 < ∣x − d∣ < δ2 with x ∈ (a, b)

implies

g(x) − g(d) > 0. x−d

This last inequality implies g(x) < g(d) if x < d, so e is not the end point d. 5.3.5 (1) (2) (3) (4) (5) (6)



Exercises

Prove the equality (5.6). Finish the proof of Theorem 5.10 by proving the case when f ′ (d) < r < f ′ (c). √ Use Taylor’s Theorem to approximate 5 to within 0.0001. Use Taylor’s Theorem to approximate cos(0.1) to within 0.0001. Use Taylor’s Theorem to approximate ln 2 to within 0.001. Prove the inequalities 1−

x2 x2 x4 ≤ cos x ≤ 1 − + 2 2 4!

x−

x3 x5 x3 ≤ sin x ≤ x − + 3! 3! 5!

for x > 0. (7) Prove the inequalities

for x > 0. (8) Prove the inequalities 2N

k−1 x

∑ (−1) k=1

k

k

for each positive integer N and (9) Prove the inequalities √ x 2+ √ − 2 2 for x > 0.

2N +1

< ln(1 + x) < ∑ (−1)k−1 k=1

xk k

each positive real number x. √ √ x2 x √ < 2+x< 2+ √ 8 2 2 2

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(10) Prove the inequality ex > 1 + x +

xn x2 +⋯+ 2! n!

for x > 0 and n ≥ 1. (11) Compute the limits. sin 4x x 9 limx→1 xx7 −1 −1 x −e limx→1 ee2x −e 2 ln x limx→1 cos( πx ) 2

(a) limx→0 (b) (c) (d)

(12) Prove the following generalization of Exercise (16) on p. 126. Suppose n ≥ 1, f has continuous (n + 1)st derivative on an open interval I, and a is a point in I such that 0 = f (a) = f ′ (a) = ⋯ = f (n) (a) and f (n+1) (a) =/ 0. Then there exists an open interval J ⊆ I containing a on which f (k) is non-zero for 0 ≤ k ≤ n + 1 with the exception of the point a. (13) Prove the following generalization of L’Hospital’s Rule. Suppose n ≥ 1, f and g both have continuous (n + 1)st derivatives on an open interval I, a is a point in I such that 0 = f (k) (a) = g (k) (a) for 0 ≤ k ≤ n, and that g (n+1) (a) =/ 0. Then f (x) f (n+1) (a) = (n+1) . x→a g(x) g (a) lim

(14) Compute the limits cos x−1 x2 2 x limx→0 tan x2 limx→0 tanxx−x 3

(a) limx→0 (b) (c)

5.4

Additional Exercises

(1) Give an example of a differentiable, uniformly continuous function f on some open interval such that f ′ is unbounded. (2) Formulate and prove a version of chain rule involving the composition of three functions. (3) Prove that f (x) = ∣x3 ∣ is differentiable on (−∞, 0) and (0, ∞), and compute its derivative. (4) Let n be a positive integer. Suppose f ∶ R → R is the function f (x) = xn if x is rational and f (x) = 0 if x is irrational. Prove that f is differentiable at 0, and compute f ′ (0). (5) Suppose f ′′ (c) exists. Prove that f (c + h) − 2f (c) + f (c − h) . h→0 h2

f ′′ (c) = lim

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Give an example of a function for which this limit exists, but f ′′ (c) does not exist. (6) A function f ∶ R → R is called odd if f (−x) = −f (x) for all x and is called even if f (−x) = f (x) for all x. Prove the following statements. (a) If f is a differentiable even function, then f ′ is odd. (b) If f is a differentiable odd function, then f ′ is even. (7) Prove Carath´ eodory’s Theorem: Suppose f is defined on an open interval I containing c. Then f is differentiable at c if and only if there exists a function g defined on I that is continuous at c and satisfies f (x) − f (c) = g(x)(x − c) for x ∈ I. When f is differentiable at c, prove that g(c) = f ′ (c). (8) The left-hand derivative of f at c is defined as the left-hand limit f (x) − f (c) , x→c x−c if it exists. The right-hand derivative is defined similarly using the righthand limit. Prove that f is differentiable at c if and only if its left-hand and right-hand derivatives both exist and are equal. (9) Suppose f is continuous on [a, b], differentiable on (a, b), f (a) = f (b) = 0, and r is a real number. Prove that there exists a point c in (a, b) such that f ′ (c) = rf (c). (10) Cauchy’s Mean Value Theorem says: Suppose f and g are both continuous on [a, b] and differentiable on (a, b). Then there exists a point c in (a, b) such that lim−

(f (b) − f (a))g ′ (c) = (g(b) − g(a))f ′ (c). Prove this theorem as follows. (a) First consider the case g(a) = g(b), and use Rolle’s Theorem. (b) Next consider the case g(a) =/ g(b). The required equality is equivalent to f ′ (c) f (b) − f (a) = . g ′ (c) g(b) − g(a) Let r denote the quotient on the right-hand side, and define h = f − rg. Now show that Rolle’s Theorem applies to h. (11) Give another proof of the Mean Value Theorem using Cauchy’s Mean Value Theorem. (12) Suppose f is defined on an open interval I such that ∣f (x) − f (y)∣ ≤ M (x − y)2 for some M > 0 and all x, y in I. (a) Prove that f is uniformly continuous on I. (b) Prove that f is differentiable on I.

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(c) Prove that f is a constant function on I. (13) Suppose f is differentiable on an open interval such that f ′ is strictly monotone. Prove that f ′ is continuous. (14) Suppose ⎧ 1 ⎪ ⎪x4 (2 + sin ( x )) if x =/ 0, f (x) = ⎨ ⎪ if x = 0. ⎪ ⎩0 Prove the following statements. (a) f achieves an absolute minimum at 0. (b) In each open interval containing 0, f ′ takes on both positive and negative values. This example shows that the converse of the first derivative test in differential calculus is false. (15) Suppose f ′′ exists on (a, b), and x < y < z are three points in (a, b) such that both f (x) and f (z) are strictly greater than f (y). Prove that there exists a point c in (a, b) such that f ′′ (c) > 0.

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Chapter 6

Integration

In this chapter we study integration, which is in some sense a reverse process of differentiation. Integrals are defined in section 6.1 using lower and upper sums. An alternative approach to integrals based on tagged partitions is given in section 6.2. Basic properties of integrals, including the Mean Value Theorem for integrals, are given 6.3. The Fundamental Theorem of Calculus and some of its important consequences are proved in section 6.4.

6.1

The Integral

In this section, we first discuss lower and upper sums associated to partitions. The integral is defined in terms of these lower and upper sums. We then prove a useful -criterion for integrability and use it to show that continuous functions are integrable.

6.1.1

Motivation

Let us briefly discuss the motivation behind the definitions below. As one learns in calculus, the definite integral is intuitively the area under the graph of a function over an interval. To define it algebraically, one uses a similar process as for derivative, which in nice cases represents the slope of the tangent. More precisely, in the (c) represents the slope of definition of the derivative, the difference quotient f (x)−f x−c a secant line that is close to the tangent at c. Taking the limit as x → c then yields the slope of the tangent at c for nice functions. The integral for nice functions can be intuitively represented as the area under the graph. To approximate it, we first cut the interval into several sub-intervals. Over each small sub-interval, we can approximate the area using a rectangle whose height is the value of the function at a chosen point. One has choices as to which points to use for the heights. Two obvious choices are the maximum and the minimum, which lead to circumscribed and inscribed rectangles. We can then define the area/integral in terms of these rectangles. Since these approximations yield sets 135

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of numbers rather than functions, we will replace limx→c in the derivative with suitable infimum and supremum. The definitions below are formal versions of these ideas. 6.1.2

Upper and Lower Sums

Throughout the rest of this chapter, unless otherwise specified, f is a bounded function defined on a closed interval [a, b]. Recall from Definition 4.9 the concept of a partition of [a, b]. Definition 6.1. Suppose P = {a = x0 < x1 < ⋯ < xn = b} is a partition of [a, b]. (1) Its norm is defined as ∣∣P ∣∣ = max{xi − xi−1 ∶ i = 1, . . . , n}. (2) Define ● ∆xi = xi − xi−1 , ● li (f ) = inf{f (x) ∶ x ∈ [xi−1 , xi ]}, and ● ui (f ) = sup{f (x) ∶ x ∈ [xi−1 , xi ]} for i = 1, . . . , n. (3) The lower sum of f with respect to P is the sum n

L(f ; P ) = ∑ li (f )∆xi . i=1

(4) The upper sum of f with respect to P is the sum n

U (f ; P ) = ∑ ui (f )∆xi . i=1

(5) If Q is another partition of [a, b] such that P ⊆ Q, then Q is called a refinement of P . The reader should be careful that the infimum li (f ) and the supremum ui (f ) may not be in the range of f . So the lower sum and the upper sum with respect to a partition are not necessarily Riemann sums in the calculus sense. Definition 6.2. The lower sum and the upper sum of f on [a, b] are defined as the least upper bound L(f ) = sup{L(f ; P ) ∶ P a partition of [a, b]} and the greatest lower bound U (f ) = inf{U (f ; P ) ∶ P a partition of [a, b]}, respectively.

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A major confusing point about the lower sum and the upper sum is that each of them involves both infimum and supremum. For example, to define the lower sum, first we form the lower sum with respect to a partition P by using the infimum li (f ) on each sub-interval of P . Then we take the supremum of the lower sums with respect to the partitions. Likewise, the upper sum involves the supremum ui (f ) on each sub-interval of a partition and then the infimum of the upper sums with respect to partitions. Notice that if we switch the order of the infimum and the supremum in the definition of the lower sum, the result is the upper sum, and vice versa. As we will see shortly below, being integrable is essentially saying that the infimum and the supremum operations commute. To define the integral, we first need to establish some basic properties of the lower sum and the upper sum. Theorem 6.1. Suppose f is a bounded function on [a, b], and P ⊆ Q are partitions of [a, b]. Then m(b − a) ≤ L(f ; P ) ≤ L(f ; Q) ≤ U (f ; Q) ≤ U (f ; P ) ≤ M (b − a), where m = inf{f (x) ∶ x ∈ [a, b]} and M = sup{f (x) ∶ x ∈ [a, b]}. Proof.

For the left-most inequality, we have m ≤ li (f ) for each i, so n

n

m(b − a) = ∑ m∆xi ≤ ∑ li (f )∆xi = L(f ; P ). i=1

i=1

For the second inequality, first note that Q has only finitely many points more than P . Thus, by an induction argument it is enough to prove the case where Q has exactly one partition point t ∈ (xi−1 , xi ) more than P , where xi−1 and xi are two consecutive partition points in P . Denote the infimum of f on [xi−1 , t] and [t, xi ] by li′ (f ) and li′′ (f ), respectively. Since ∆xi = (xi − t) + (t − xi−1 ), we have li (f )∆xi = li (f )(t − xi−1 ) + li (f )(xi − t) ≤ li′ (f )(t − xi−1 ) + li′′ (f )(xi − t) because li (f ) ≤ li′ (f ) and li (f ) ≤ li′′ (f ). Now on [a, xi−1 ] and [xi , b], the partitions P and Q have the same partition points, so the previous inequality implies the desired inequality L(f ; P ) ≤ L(f ; Q). The third inequality holds because li (f ) ≤ ui (f ) for each i. The other two inequalities are left as exercises for the reader.  The previous result actually implies that each lower sum is no more than each upper sum with respect to arbitrary partitions. Intuitively, a lower sum with respect

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to some partition consists of inscribed rectangles, so the sum of their areas cannot exceed that of circumscribed rectangles with respect to any other partition. Corollary 6.1. Suppose f is a bounded function on [a, b], and P and Q are two partitions of [a, b]. Then L(f ; P ) ≤ U (f ; Q). Proof. yields

The union P ∪Q is a refinement of both P and Q, so the previous theorem L(f ; P ) ≤ L(f ; P ∪ Q) ≤ U (f ; P ∪ Q) ≤ U (f ; Q),

(6.1)

as desired.



Corollary 6.2. Suppose f is a bounded function on [a, b], and P and Q are two partitions of [a, b]. Then L(f ; P ) ≤ L(f ) ≤ U (f ) ≤ U (f ; Q). Proof. The two outer inequalities hold by the definitions of the lower sum and the upper sum. For the middle inequality, we know that L(f ; P ) is a lower bound of every U (f ; Q) by Corollary 6.1. Since U (f ) is the greatest lower bound of these U (f ; Q), we have L(f ; P ) ≤ U (f ). This inequality says that U (f ) is an upper bound of L(f ; P ). Since P is arbitrary, the middle inequality follows.  6.1.3

Integrability

Definition 6.3. A bounded function f on [a, b] is said to be integrable on [a, b] if its lower sum and upper sum are equal, that is, L(f ) = U (f ). In this case, the common value is denoted by b

∫

b

f

or

a

f (x)dx,

∫

a

and it is called the integral of f on [a, b]. If f is integrable on [a, b], we define a

∫

b

b

f = −∫

f.

a

From now on, when we say f is integrable on [a, b], we automatically assume that f is bounded on [a, b]. When f is integrable on [a, b], the dummy variable x in b ∫a f (x)dx may be changed to any other symbol, such as u. We will discuss several classes of integrable functions. Before that, let us first exhibit a function that is not integrable.

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Example 6.1. Here we observe that the characteristic function χQ of Q defined in (4.2) is not integrable on any interval [a, b]. Indeed, if P is a partition of [a, b], then in each sub-interval [xi−1 , xi ] there are both rational and irrational numbers. So li (χQ ) = 0

and

ui (χQ ) = 1

for each i. This implies L(χQ ; P ) = 0

and

U (χQ ; P ) = ∑ ∆xi = b − a.

Since P is an arbitrary partition, we have L(χQ ) = 0

and

U (χQ ) = b − a,

which shows that χQ is not integrable on [a, b]. 6.1.4

Criterion for Integrability

To establish integrability of functions, the following -criterion is often useful. Theorem 6.2. A bounded function f on [a, b] is integrable if and only if, for each  > 0, there exists a partition P of [a, b] such that U (f ; P ) − L(f ; P ) < . Proof. The “only if” part is an 2 -argument. Suppose f is integrable on [a, b], so b I = ∫a f = L(f ) = U (f ). Given  > 0, we have   I − < L(f ) and I + > U (f ), 2 2 so there exist partitions P1 and P2 such that   L(f ; P1 ) > I − and U (f ; P2 ) < I + . (6.2) 2 2 If P = P1 ∪ P2 , then (6.1) and (6.2) imply   U (f ; P ) − L(f ; P ) ≤ U (f ; P2 ) − L(f ; P1 ) < + = , 2 2 proving the “only if” part. For the “if” part, suppose  > 0 is given, and P is a partition as in the statement of the theorem. Then Corollary 6.2 implies 0 ≤ U (f ) − L(f ) ≤ U (f ; P ) − L(f ; P ) < . Since  is arbitrary, we conclude that U (f ) = L(f ).



The meaning of the previous theorem is that, a function is integrable exactly when the lower sum and the upper sum can be made arbitrarily close to each other by picking suitable partitions. This theorem is often easier to use than the original definition because it involves finding only one suitable partition. In the original definition of integrability, one has to consider all the partitions of the interval.

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Integrability of Continuous Functions

We now illustrate the above -criterion by showing that continuous functions are integrable. Theorem 6.3. Suppose f is a continuous function on [a, b]. Then f is integrable on [a, b]. Ideas of Proof. We need to estimate n

U (f ; P ) − L(f ; P ) = ∑ (ui (f ) − li (f )) ∆xi . i=1

The plan is to choose a partition P with sufficiently small norm such that each difference (ui (f ) − li (f )) is suitably small. The sum of the ∆xi is, in any case, (b − a). Proof. Suppose given  > 0. The function f is uniformly continuous on [a, b] by  , there exists δ > 0 such that ∣x − y∣ < δ with Theorem 4.9. Thus, given 0 = b−a x, y ∈ [a, b] implies  . ∣f (x) − f (y)∣ < 0 = b−a Pick any partition P of [a, b] with ∣∣P ∣∣ < δ; Exercise (3) below guarantees its existence. By the Extreme Value Theorem 4.6, both ui (f ) and li (f ) are in the range of f on [xi−1 , xi ], so ui (f ) − li (f ) < 0 for each i. Thus, we have n

U (f ; P ) − L(f ; P ) < ∑ 0 ∆xi = 0 (b − a) = . i=1

Theorem 6.2 now says that f is integrable on [a, b].



In the exercises and later sections, we will see that a general integrable function may have many points of discontinuity. 6.1.6

Exercises

(1) If P is a partition of [a, b], prove that ∆xi ≤ ∣∣P ∣∣ for each i and that ∑ni=1 ∆xi = b − a. (2) Suppose P ⊆ Q are partitions of [a, b]. Prove that ∣∣Q∣∣ ≤ ∣∣P ∣∣ ≤ b − a. (3) Suppose [a, b] is a closed interval, and δ > 0. Prove that there exists a partition P of [a, b] with ∣∣P ∣∣ < δ. (4) Using Theorem 6.2, write down what it means for a bounded function f on [a, b] to not be integrable on [a, b].

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(5) Suppose f is integrable on [a, b], and there exists a real number r such that L(f ; P ) ≤ r ≤ U (f ; P ) b

for every partition P of [a, b]. Prove that r = ∫a f . (6) Suppose f and g are bounded functions on [a, b], and P is a partition on [a, b]. Prove that ui (f + g) ≤ ui (f ) + ui (g)

and li (f + g) ≥ li (f ) + li (g)

for each i. (7) In the proof of the second inequality in Theorem 6.1, an induction argument was used. Write down the details of this induction. (8) Prove the last two inequalities in Theorem 6.1. (9) Prove that every constant function is integrable on every closed bounded interval. Write down an expression for the integral. (10) Prove that the step function J in Exercise (10) on page 104 is integrable on every closed bounded interval. (11) Suppose f is an increasing function on [a, b], and P is a partition of [a, b]. Prove that U (f ; P ) − L(f ; P ) ≤ ∣∣P ∣∣(f (b) − f (a)). (12) Using the previous exercise or otherwise, prove that an increasing function on [a, b] is integrable. (13) Prove that every monotone function is integrable on [a, b]. (14) Suppose n is a positive integer, and [ai , bi ] ⊆ [a, b] for i = 1, . . . , n are n subintervals such that bi ≤ ai+1 for 1 ≤ i ≤ n − 1. Prove that n

∑ [U (f ; [ai , bi ]) − L(f ; [ai , bi ])] ≤ (M − m)(b − a), i=1

where m = inf{f (x) ∶ x ∈ [a, b]} and M = sup{f (x) ∶ x ∈ [a, b]}. (15) Using the previous exercise or otherwise, prove the following Cauchy criterion for integrability. A bounded function f on [a, b] is integrable on [a, b] if and only if, for each  > 0, there exists δ > 0 such that U (f ; P ) − L(f ; P ) <  for every partition P of [a, b] with ∣∣P ∣∣ < δ.

6.2

Integration via Tagged Partitions

There is another rigorous approach to integrals based on Riemann sums that is much closer to the usual presentation in calculus. In this section we discuss integrability based on Riemann sums. The main idea of Riemann sums is that, instead of taking the infimum or the supremum of f on each sub-interval, one takes the value of f at some point. Intuitively, this process forms a rectangle whose height lies between li (f ) and ui (f ).

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A First Course in Analysis

Riemann Sums

To define Riemann sums, first we need to define partitions with chosen points in each sub-interval. Definition 6.4. A tagged partition P = {{xj }nj=0 , {wi }ni=1 } of an interval [a, b] consists of: ● a partition {x0 < ⋯ < xn } of [a, b] and ● a point wi in [xi−1 , xi ] for each i = 1, . . . , n. The norm of such a tagged partition is defined as ∣∣P ∣∣ = max{xi − xi−1 ∶ i = 1, . . . , n}. Given a tagged partition P , one can forget about the points wi and obtain a partition, which we also denote by P . Conversely, given a partition P of [a, b], there are many different tagged partitions with the same partition points xj that can be associated to it, which we denote by P ′ , P ′′ , etc. Definition 6.5. Given a bounded function f on [a, b] and a tagged partition P = {{xj }nj=0 , {wi }ni=1 } of [a, b], the Riemann sum of f with respect to P is the sum n

R(f ; P ) = ∑ f (wi )∆xi , i=1

where ∆xi = xi − xi−1 as in Definition 6.1 Whenever we use the symbol R(f ; P ), it is automatically assumed that P is a tagged partition. If the lower sum L(f ; P ) or the upper sum U (f ; P ) are also considered, then we are using the partition associated to the tagged partition P . The following basic observation tells us how the lower sum, the upper sum, and the Riemann sum are related. The proof of this observation is left as an exercise. Lemma 6.1. Suppose f is a bounded function on [a, b], and P is a tagged partition of [a, b]. Then L(f ; P ) ≤ R(f ; P ) ≤ U (f ; P ). Therefore, for a given tagged partition, the Riemann sum always lies between the lower sum and the upper sum. Intuitively, this is true because in each sub-interval, a typical rectangle whose height is a value of f must lie between the inscribed and the circumscribed rectangles. The next observation says that the lower sum and the upper sum can be closely approximated by Riemann sums.

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Theorem 6.4. Suppose f is a bounded function on [a, b], P is a partition of [a, b], and  > 0. Then there exist tagged partitions P ′ and P ′′ with the same partition points as P such that R(f ; P ′ ) − L(f ; P ) < 

U (f ; P ) − R(f ; P ′′ ) < .

and

Ideas of Proof. The lower sum is formed using the infimum of f on each subinterval. To construct the tagged partition P ′ , in each sub-interval we choose a point whose value under f is very close to the infimum. The resulting Riemann sum is then close to the lower sum. Proof.

Set 0 =

 . b−a

If P = {x0 < ⋯ < xn }, then since li (f ) + 0 > li (f )

for each i, there exists a point wi in [xi−1 , xi ] such that li (f ) ≤ f (wi ) < li (f ) + 0 . So we have 0 ≤ f (wi ) − li (f ) < 0 . Let P ′ be the tagged partition with chosen points wi and the same partition points as P . Multiplying by ∆xi and summing over the sub-intervals, we obtain the inequalities n

n

R(f ; P ′ ) − L(f ; P ) = ∑[f (wi ) − li (f )]∆xi < 0 ∑ ∆xi = 0 (b − a) = . i=1

i=1

This proves the first inequality. The proof of the other inequality is left as an exercise.  6.2.2

Riemann Integrability

We now define a version of integrability based on Riemann sums. Definition 6.6. Suppose f is a bounded function on [a, b]. Then f is said to be b Riemann integrable on [a, b] if there exists a real number R ∫a f such that, for each  > 0, there exists δ > 0 such that b

∣R(f ; P ) − R ∫

f∣ < 

a b

for every tagged partition P with ∣∣P ∣∣ < δ. In this case, the real number R ∫a f is called the Riemann integral of f on [a, b].

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Note that we are using the notation R ∫a f to distinguish it from the integral in Definition 6.3 because at this moment we have not shown that they are equal. We also have not shown that integrability is equivalent to Riemann integrability. We now show that the two approaches to integration are equivalent. b ∫a f

Theorem 6.5. A bounded function f on [a, b] is integrable if and only if it is b b Riemann integrable. In this case, the integral ∫a f and the Riemann integral R ∫a f are equal. b

Ideas of Proof. Under the assumption of integrability, the integral ∫a f can be closely approximated by lower and upper sums, which in turn can be closely approximated by Riemann sums. Likewise, with Riemann integrability, the Riemann b integral R ∫a f can also be closely approximated by Riemann sums. Therefore, it should be intuitively clear that the two approaches are equivalent. Also, it should not be surprising that some sort of 2 -argument is involved as we link the integral to the Riemann integral via lower, upper, and Riemann sums. Proof. For the “only if” part, we assume f is integrable on [a, b], and suppose  > 0. By Exercise (15) on page 141, there exists δ > 0 such that U (f ; P ) − L(f ; P ) < 

(6.3)

for every partition P with ∣∣P ∣∣ < δ. Now if P is any tagged partition, then b

L(f ; P ) ≤ R(f ; P ) ≤ U (f ; P )

and L(f ; P ) ≤ ∫

f ≤ U (f ; P )

a

by Lemma 6.1 and integrability. Therefore, if P has norm < δ, then (6.3) implies b

∣R(f ; P ) − ∫

f ∣ < .

a b

b

This shows that f is Riemann integrable and ∫a f = R ∫a f . For the “if” part, we assume f is Riemann integrable. We will use the -criterion in Theorem 6.2 and an 4 -argument to show that f is integrable. Given  > 0, Riemann integrability implies that there exists δ > 0 such that b

 4 for every tagged partition P with norm < δ. Pick any partition P with norm < δ. By Theorem 6.4, there exist tagged partitions P ′ and P ′′ with the same partition points as P , and hence with norms < δ, such that   R(f ; P ′ ) − L(f ; P ) < and U (f ; P ) − R(f ; P ′′ ) < . 4 4 Since we can write ∣R(f ; P ) − R ∫

f∣ <

a

b

U (f ; P ) − L(f ; P ) = [U (f ; P ) − R(f ; P ′′ )] + [R(f ; P ′′ ) − R ∫

f]

a

b

+ [R ∫

a

f − R(f ; P ′ )] + [R(f ; P ′ ) − L(f ; P )],

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the previous three inequalities and the Triangle Inequality imply  (6.4) ∣U (f ; P ) − L(f ; P )∣ < 4 ⋅ = . 4 Theorem 6.2 now says that f is integrable. To see that the integral is equal to the Riemann integral in this case, we use the same inequalities in the previous paragraph as follows. b   < R∫ f + 4 2 a 3 ′ < R(f ; P ) + < L(f ; P ) +  ≤ U (f ; P ) + . 4

U (f ; P ) < R(f ; P ′′ ) +

Therefore, we have b

∣R ∫

a

 f − U (f ; P )∣ < . 2

But we also have b

∣U (f ; P ) − ∫

f∣ < 

a

by (6.4). Combining the previous two inequalities, we obtain b

∣R ∫

a

b

f −∫

a

b

f ∣ ≤ ∣R ∫

a

b

f − U (f ; P )∣ + ∣U (f ; P ) − ∫

a

b

b

Since  > 0 is arbitrary, we conclude that R ∫a f = ∫a f .

f∣ <

3 2 

In view of Theorem 6.5, the words integrable and Riemann integrable are interchangeable. 6.2.3

Exercise

(1) Using Definition 6.6, write down what it means for a bounded function f on [a, b] to not be Riemann integrable on [a, b]. (2) Prove Lemma 6.1. (3) Prove the second inequality in Theorem 6.4. (4) Without using Theorem 6.5, prove that when f is Riemann integrable on [a, b], b the value of the Riemann integral R ∫a f is unique. (5) Prove the following Cauchy criterion for Riemann integrability. A bounded function f on [a, b] is Riemann integrable if and only if, for every  > 0, there exists δ > 0 such that ∣R(f ; P ) − R(f ; Q)∣ <  for any two tagged partitions P and Q with norms < δ. (6) Using the previous exercise, prove that the characteristic function χQ of Q is not Riemann integrable on [0, 1].

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(7) Suppose f is Riemann integrable on [a, b] and that each Pn is a tagged partition with lim ∣∣Pn ∣∣ = 0. Prove that b

f = lim R(f ; Pn ).

R∫

n→∞

a

(8) For an integrable function f on [0, 1], prove that 1

∫

0

n n i 1 i−1 1 f = lim ∑ f ( ) = lim ∑ f ( ) . n→∞ n→∞ n n n n i=1 i=1

(9) Give an example of a bounded function f for which the limits in the previous exercise both exist, but f is not integrable on [0, 1]. 1 (10) Give an example of an integrable function f on [0, 1] such that ∫0 f = 0 and that f (x) =/ 0 for all x in [0, 1]. (11) Prove that the function f ∶ [0, 1] → R given by f (x) = sin ( x1 ) if x =/ 0 and f (0) = 0 is integrable. (12) Consider the function f ∶ R → R given by f (x) = x if x is rational and f (x) = −x if x is irrational. Prove that f is not integrable on any closed bounded interval. 6.3

Basic Properties of Integrals

In this section we establish some basic properties of the integral, most of which should already be familiar to the reader from calculus. 6.3.1

Linearity

The next result says that integrals are linear and respect inequalities. Theorem 6.6. Suppose f and g are integrable on [a, b], and r is a real number. (1) The sum f + g is integrable on [a, b], and b

b

b

∫ (f + g) = ∫ a

f +∫

a

g.

a

(2) The scalar multiple rf is integrable on [a, b], and b

b

rf = r ∫

∫

a

f.

a

(3) If f (x) ≥ 0 for all x in [a, b], then b

∫

f ≥ 0.

a

Proof. We will prove the first part. The other two parts are left as exercises. Suppose P1 and P2 are partitions of [a, b] and P = P1 ∪P2 . Then we have inequalities L(f ; P1 ) + L(g; P2 ) ≤ L(f ; P ) + L(g; P ) ≤ L(f + g; P ) ≤ U (f + g; P ) ≤ U (f ; P ) + U (g; P ) ≤ U (f ; P1 ) + U (g; P2 ).

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The first, the third, and the fifth inequalities follow from Theorem 6.1, while the second and the fourth inequalities follow from Exercise (6) on page 141. Using the above inequalities and Corollary 6.2, we obtain the inequalities L(f ; P1 ) + L(g; P2 ) ≤ L(f + g) ≤ U (f + g) ≤ U (f ; P1 ) + U (g; P2 ). Since this holds for arbitrary partitions P1 and P2 , it follows that b

∫

a

b

f +∫

a

b

g ≤ L(f + g) ≤ U (f + g) ≤ ∫

a

b

f +∫

g,

a

from which the first part follows.



Corollary 6.3. Suppose f and g are integrable on [a, b] such that f (x) ≥ g(x) for all x in [a, b]. Then b

∫

a

Proof.

b

f ≥∫

g.

a

By Theorem 6.6, the difference f − g is integrable, and b

b

0 ≤ ∫ (f − g) = ∫ a

a

b

f −∫

g,

a

from which the desired inequality follows. 6.3.2



Consecutive Intervals

The next result says that integrals can be added over two consecutive intervals. Theorem 6.7. Suppose f is integrable on [a, c] and on [c, b]. Then f is integrable on [a, b] and b

∫

a

c

f =∫

a

b

f +∫

f.

c

Ideas of Proof. Partitions of [a, c] and [c, b] can be spliced together to form a partition of [a, b]. This implies that the lower and the upper sums also have the same property. Therefore, close estimates between the lower and the upper sums on [a, c] and [c, b] can be extended to all of [a, b]. Proof. This is an 2 -argument. Given  > 0, there exist partitions P1 of [a, c] and P2 of [c, b] such that  U (f ; Pi ) − L(f ; Pi ) < 2 for each i. The union P = P1 ∪ P2 is a partition of [a, b], and 2

U (f ; P ) − L(f ; P ) = ∑(U (f ; Pi ) − L(f ; Pi )) < i=1

  + = . 2 2

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Therefore, f is integrable on [a, b]. To prove the desired equality, consider the inequalities c

∫

a

b

f +∫

c

f −  ≤ U (f ; P1 ) + U (f ; P2 ) −  = U (f ; P ) −  b

< L(f ; P ) ≤ ∫

f ≤ U (f ; P ) < L(f ; P ) + 

a c

= L(f ; P1 ) + L(f ; P2 ) +  ≤ ∫

a

b

c

b

f +∫

f + .

c

b

The desired equality ∫a f = ∫a f + ∫c f now follows because  is arbitrary. a ∫b f



b − ∫a f

Since we defined = if f is integrable on [a, b], the previous theorem is actually valid for all a, b, and c, as long as the three integrals exist. 6.3.3

Mean Value Theorem for Integrals

Intuitively, this result says that the integral of a continuous function is actually the area of a rectangle, whose width is that of the interval and whose height is the value of the function at a certain point. Theorem 6.8. Suppose f is continuous on [a, b]. Then there exists a point c in [a, b] such that b

∫

f = f (c)(b − a).

a

Proof. Theorem 6.3 says that f is integrable. The result is obvious if f is a constant function, so we assume that f is not constant. We are trying to show that the real number b 1 r= ∫ f b−a a is in the range of f . By the Extreme Value Theorem 4.6, f achieves its maximum and minimum on [a, b] at some points β and α. If r is either f (α) or f (β), then we are done. Otherwise, by the Intermediate Value Theorem 4.7 applied to the interval with end points α and β, it is enough to observe that r lies between f (α) and f (β). From Theorem 6.1 we know that b

f (α)(b − a) ≤ ∫

f ≤ f (β)(b − a),

a

from which we obtain the desired inequalities f (α) ≤ r ≤ f (β). 6.3.4

Exercises

(1) Finish the proof of Theorem 6.6 by proving the last two parts. (2) Suppose f is integrable on [a, b]. Prove that ∣f ∣ is integrable on [a, b] and b

∣∫

a

b

f∣ ≤ ∫

a

∣f ∣.



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(3) Give an example where ∣f ∣ is integrable, but f is not. (4) Suppose f is integrable on [a, b] and [c, d] ⊆ [a, b]. Prove that f is integrable on [c, d]. (5) Suppose f is integrable on [a, b], g is a bounded function on [a, b], and that f (x) = g(x) on [a, b] except for finitely many points. Prove that g is integrable b b on [a, b] and ∫a f = ∫a g. b (6) Suppose f is continuous on [a, b] such that ∫a f = 0. Prove that f (c) = 0 for some point c in [a, b]. b b (7) Suppose f and g are continuous on [a, b] such that ∫a f = ∫a g. Prove that f (c) = g(c) for some point c in [a, b]. b (8) Suppose f is continuous on [a, b] such that f (x) ≥ 0 for all x and ∫a f = 0. Prove that f (x) = 0 for all x. (9) Suppose f is integrable on [a, b], g is continuous on [c, d], and that f ([a, b]) ⊆ [c, d]. Prove that the composition g ○ f is integrable on [a, b]. (10) Suppose f is integrable on [a, b]. Prove that the product f 2 is integrable on [a, b]. (11) Suppose f and g are integrable on [a, b]. Prove that the product f g is integrable on [a, b]. (12) Suppose f and g are continuous on [a, b] and g(x) > 0 for all x. Prove that there exists a point c in [a, b] such that b

∫

a

b

f g = f (c) ∫

g.

a

Use this result to give another proof of Theorem 6.8. (13) Give an example to show that in Theorem 6.8, the continuity assumption cannot be weakened to integrability. (14) Consider the function on [0, 1] given by f (x) = 0 if x is irrational, f (0) = 1, and f ( m ) = n1 if m is a rational number in lowest terms. Prove the following n n statements. (a) f is continuous at each irrational number. (b) f is discontinuous at each rational number. (c) f is integrable. This function is called the Thomae function.

6.4

Fundamental Theorem of Calculus

The main purpose of this section is to prove the theorem in the section title, which the reader should already be familiar with from calculus. Some important consequences are then discussed. As in the previous sections, our goal here is to prove these theorems rigorously, rather than doing examples about how to use them as in calculus.

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Motivation

Just like chain rule is the most important derivative rule in differential calculus, the Fundamental Theorem of Calculus is the most important result in integral calculus. Chain rule is important because a lot of functions are compositions of simpler functions. Computing the actual value of an integral is, however, an entirely different matter. It involves first computing the lower and upper sums with respect to every partition of the interval, and then taking the supremum of the lower sums and the infimum of the upper sums. Moreover, the integrals of the constituent functions do not usually tell us what the integral is for the composition. For example, try to 1 1 1 compute ∫0 x2 , and see if it helps you compute ∫0 x4 and ∫0 x6 from the definition. It is therefore necessary to develop a more convenient method for computing the integral than using the definition. The Fundamental Theorem of Calculus serves exactly this purpose. It drastically simplifies the computation of the integral for a function that is actually the derivative of another function. 6.4.2

Main Theorem

Here is the Fundamental Theorem of Calculus. Theorem 6.9. Suppose f is continuous on [a, b], differentiable on (a, b), and that f ′ is integrable on [a, b]. Then b

f ′ = f (b) − f (a).

∫

a

Ideas of Proof. We need to estimate b

∣∫

f ′ − (f (b) − f (a))∣ .

a

b

Since ∫a f ′ can be closely estimated by the lower and the upper sums, we will try to do the same to the difference (f (b) − f (a)). Remember the Mean Value Theorem 5.7 relates such a difference to the derivative, so it is no surprise that we will use this theorem. Proof. Given  > 0, by Theorem 6.2, there exists a partition P = {x0 < ⋯ < xn } of [a, b] such that U (f ′ ; P ) − L(f ′ ; P ) < . Since  is arbitrary and b

L(f ′ ; P ) ≤ ∫

f ′ ≤ U (f ′ ; P ),

a

it is enough to show that (f (b) − f (a)) satisfies the same inequalities, that is, L(f ′ ; P ) ≤ f (b) − f (a) ≤ U (f ′ ; P ).

(6.5)

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Applying the Mean Value Theorem to each sub-interval [xi−1 , xi ], we obtain a point wi in (xi−1 , xi ) such that f ′ (wi )∆xi = f (xi ) − f (xi−1 ). Since li (f ′ ; P ) ≤ f ′ (wi ) ≤ ui (f ′ ; P ), we conclude that n

n

L(f ′ ; P ) = ∑ li (f ′ ; P )∆xi ≤ ∑(f (xi ) − f (xi−1 )) i=1

i=1 n

≤ ∑ ui (f ′ ; P )∆xi = U (f ′ ; P ). i=1

The inequalities (6.5) now follow because n

∑(f (xi ) − f (xi−1 )) = f (b) − f (a). i=1



Using the Fundamental Theorem of Calculus to compute integrals is something the reader should already know from calculus. We adopt the notation b

f (b) − f (a) = f ∣ a

from calculus. We now consider other important integration theorems. 6.4.3

Integration by Parts

This theorem is the integral form of the product rule (Theorem 5.2), which says (f g)′ = f ′ g + f g ′ . Roughly speaking, integrating both sides from here yields Integration by Parts. The formal proof uses the Fundamental Theorem of Calculus. Corollary 6.4. Suppose f and g are continuous on [a, b], differentiable on (a, b), and that f ′ and g ′ are integrable on [a, b]. Then b

∫

a

b

b

f g′ = f g ∣ − ∫ a

f ′ g.

a

Proof. The desired equality is exactly that in the Fundamental Theorem of Calculus 6.9 for the product h = f g, which satisfies the hypotheses by Exercise (3) on page 93, Theorem 5.2, Theorem 6.3, and Exercise (11) on page 149. 

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Integral Form of Taylor’s Theorem

In Taylor’s Theorem 5.8, it was shown that for n ≥ 1 the error term Rn (x; a) = f (x) − Tn (x; a) in the degree n Taylor polynomial approximation of f at a takes the form f (n+1) (c) (x − a)n+1 (n + 1)! for some point c between a and x. We now present another form of this error term that does not give preference to any particular point in the interval determined by a and x. To do that, we will have to use the integral. This result is also a good demonstration of the Fundamental Theorem of Calculus and Integration by Parts. We want to keep using x as the independent variable in the integral, so the point x in the error term will be called b below. Theorem 6.10. Suppose f ∶ I = [α, β] → R is a function such that f (k) exists on (α, β) and is continuous on I for 0 ≤ k ≤ n + 1. Then given two distinct points a and b in (α, β), we have Rn (b; a) = Proof.

b 1 n (n+1) (x)dx. ∫ (b − x) f n! a

This is an induction proof for k = 1, 2, . . . , n. The kth case is Rk (b; a) =

b 1 k (k+1) , ∫ (b − x) f k! a

(6.6)

so the first case is the assertion b

R1 (b; a) = f (b) − [f (a) + f ′ (a)(b − a)] = ∫ (b − x)f ′′ .

(6.7)

a

To prove the first case, we use the Fundamental Theorem of Calculus 6.9 and Integration by Parts with g = x − b. Since g ′ = 1, the middle expression in (6.7) can be rewritten as b

b

b

b

′′ ′′ ′ ′ ′ ′ ∫ (f − f (a)) = (f − f (a))g ∣a − ∫ (x − b)f = ∫ (b − x)f a

a

a

because (f ′ − f ′ (a))(a) = 0 = g(b). This proves the first case of (6.6). The induction step is a similar argument using Integration by Parts, which is left as an exercise. 

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Derivative of the Integral

In many calculus textbooks, the next two results are often collectively called the first Fundamental Theorem of Calculus. The theme here is that a function defined by an integral is a little bit nicer than the function being integrated. Theorem 6.11. Suppose f is integrable on [a, b]. Define a function on [a, b] by x

F (x) = ∫

f.

a

Then F is uniformly continuous on [a, b]. Ideas of Proof. The function F can be intuitively interpreted as the area under f over [a, x]. As x moves within the interval [a, b], the values of F vary nicely because f is integrable. For the next two proofs, we need to estimate x

F (x) − F (y) = ∫

y

f −∫

a

a

x

f =∫

f.

y

We can do this by estimating f and ∣x − y∣. Proof.

Since f is bounded, there exists M > 0 such that ∣f (x)∣ ≤ M

Given  > 0, let δ =

 . M

for all

x ∈ [a, b].

Then for x > y in [a, b] with x − y < δ, we have x

∣F (x) − F (y)∣ ≤ ∫

∣f ∣ ≤ M (x − y) < .

y

We have a similar inequality when x < y. This shows that F is uniformly continuous on [a, b].  Theorem 6.12. Suppose f and F are as in Theorem 6.11 and that f is continuous at a point c in (a, b). Then F is differentiable at c and F ′ (c) = f (c). Proof.

First observe that

x 1 ∫ f (c) x−c c c for x =/ c. Given  > 0, the continuity of f at c implies that there exists δ > 0 such that x

F (x) − F (c) = ∫

f

and f (c) =

∣x − c∣ < δ with x ∈ [a, b] implies ∣f (x) − f (c)∣ < . Therefore, for such x we have ∣

x F (x) − F (c) 1 − f (c)∣ = ∣ ∫ (f − f (c))∣ < , x−c x−c c

which proves F ′ (c) = f (c) by Theorem 4.3.



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6.4.6

Substitution

The next result is often called u-substitution or change of variables in calculus textbooks. Theorem 6.13. Suppose g is a continuously differentiable function on an open interval I, J is an open interval containing the image of g, and f is a continuous function on J. Then for any two points a < b in I, we have b

∫

a

g(b)

f (g(x))g ′ (x)dx = ∫

f (u)du

g(a)

Proof. First note that the composition f (g(x)) is continuous on I, and therefore so is the product f (g(x))g ′ (x). It is integrable on [a, b] by Theorem 6.3. Define x

F (x) = ∫

f

g(a)

for x in g([a, b]). It is differentiable with F ′ (c) = f (c) by Theorem 6.12. Chain rule yields [F (g(x))]′ = F ′ (g(x))g ′ (x) = f (g(x))g ′ (x). Therefore, it follows from Theorem 6.9 that b

b

∫

a

g(b)

b

f (g(x))g ′ (x)dx = ∫ (F ○ g)′ = F ○ g ∣ = ∫ a

a

f,

g(a)

as desired. 6.4.7



Exercises x

b

(1) Suppose f is continuous on [a, b] such that ∫a f = ∫x f for all x in [a, b]. Prove that f is the constant function 0. x+1 (2) For a continuous function f on R, define g(x) = ∫x−1 f . Prove that g is differentiable on R. Then compute g ′ . (3) Write down the details of the proof of Corollary 6.4. (4) Finish the induction step in the proof of Theorem 6.10 as follows. (a) First prove that Rk+1 (b; a) = Rk (b; a) −

f (k+1) (a) (b − a)k+1 . (k + 1)!

(b) Suppose the kth case (6.6) is true. Prove that Rk+1 (b; a) =

b 1 (k+1) − f (k+1) (a)](b − x)k . ∫ [f k! a

(c) Apply Integration by Parts to the previous part to prove the (k + 1)st case of (6.6).

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(5) Under the hypotheses of Theorem 6.10, suppose further that a < b and that m ≤ ∣f (n+1) ∣ ≤ M on (α, β). Prove that m(b − a)n+1 M (b − a)n+1 ≤ Rn (b; a) ≤ . (n + 1)! (n + 1)! (6) Using Theorem 6.10 with a < b, give another proof of the original Taylor’s Theorem 5.8, that is, f (n+1) (c) (b − a)n+1 Rn (b; a) = (n + 1)! for some point c with a < c < b. (7) Write down the details of the proof of Theorem 6.11 in the case x < y. (8) In the proof of Theorem 6.12, we used the -δ characterization of the limit (c) limx→c F (x)−F . Write down a version of that proof that uses the sequential x−c definition of limits (Definition 4.2).

6.5

Additional Exercises

(1) Suppose f is continuous on [0, ∞), f (x) > 0 for all x > 0, and that [f (x)]n = x n ∫0 f n−1 for some integer n ≥ 2. Prove that f (x) = x for all x. Here f n−1 is the product of n − 1 copies of f . (2) Suppose f and g are integrable on [a, b]. Prove that min{f, g} and max{f, g} are both integrable on [a, b]. (3) Prove that every function of bounded variation is integrable on a closed bounded interval. (4) Suppose f and g are integrable on some intervals, that they are both monotone, and that the composition f ○ g is defined. Prove that f ○ g is integrable on the domain of g. (5) Prove that the function f (x) = 1 if x = n1 for n ∈ Z+ and f (x) = 0 otherwise is integrable on [0, 1]. (6) A function f on [a, b] is called piecewise continuous if there exists a partition P = {x0 < ⋯ < xn } of [a, b] such that f is uniformly continuous on each open sub-interval (xi−1 , xi ). Prove the following statements. (a) A piecewise continuous function on [a, b] is bounded. (b) A piecewise continuous function on [a, b] is integrable on [a, b]. (7) A function f on [a, b] is called piecewise monotone if there exists a partition P = {x0 < ⋯ < xn } of [a, b] such that f is monotone on each open sub-interval (xi−1 , xi ). Prove that a bounded piecewise monotone function on [a, b] is integrable on [a, b]. (8) Suppose f is bounded on [a, b]. Prove that f is integrable on [a, b] if and only if it is integrable on each closed sub-interval of (a, b). In this case, find b an expression of ∫a f in terms of the integrals of f over closed sub-intervals of (a, b).

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(9) Suppose f is a bounded function on [a, b] such that {R(f ; Pn )} is a convergent sequence whenever {Pn } is a sequence of tagged partitions with lim ∣∣Pn ∣∣ = 0. Prove that f is integrable on [a, b]. (10) If n is a positive integer, give an example of an integrable function on some closed bounded interval that has exactly n points of discontinuity. (11) Give an example of an increasing integrable function on some closed bounded interval that has countably many points of discontinuity. (12) Give an example of an integrable function f on [a, b] and a integrable function g on [c, d] such that f ([a, b]) ⊆ [c, d] and that the composition g ○ f is not integrable on [a, b]. (13) Suppose f is defined on [a, ∞) and is integrable on [a, b] for all b > a. Define ∞ b the improper integral ∫a f = limb→∞ ∫a f , if the limit exists. In this case, we say the improper integral converges. Prove the Integral Test as follows. Suppose ∑ an is a series with {an } decreasing and non-negative. Suppose f is a non-negative decreasing function on [1, ∞) such that f (n) = an for each n. (a) Prove that n

n

∑ ai = L(f ; P ) ≤ ∫ i=2

1

n−1

f ≤ U (f ; P ) = ∑ ai i=1

for the partition P = {1 < 2 < ⋯ < n} of [1, n]. (b) Prove that the series ∑ an converges if and only if the improper integral ∞ ∫1 f converges.

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Chapter 7

Sequences and Series of Functions

In this chapter, we discuss sequences and series of functions and their convergence behavior. One main reason for studying sequences of functions is that it is a very convenient way to construct examples of functions with desired properties. Two concepts of convergence, pointwise and uniform, of sequences of functions are introduced in section 7.1. In section 7.2 we discuss properties that are sometimes preserved by the limits of a sequence of functions. Series of functions, which are a particular kind of sequences of functions, are discussed in section 7.3. Power series, which are a particular kind of series of functions with very nice properties, are discussed in section 7.4.

7.1

Pointwise and Uniform Convergence

In this section, we first introduce pointwise convergence of a sequence of functions. It is the most straight forward way to define convergence in this setting. Several examples are then given to illustrate the inadequacy of this form of convergence. Then we introduce a stronger form of convergence called uniform convergence. In the next section, we will show that uniform convergence allows the passage of properties to the limits. 7.1.1

Pointwise Convergence

Throughout this section, unless otherwise specified, {fn } denotes a sequence of functions f1 , f2 , . . ., all having the same domain S ⊆ R. Unlike a sequence of real numbers as in Definition 2.1, there are now three ways to consider limits of {fn (x)}. First, we can consider limits with respect to the sequential index n for a fixed point x in S. Second, we can consider limits of fn (x) with respect to the points x ∈ S for a fixed index n. Finally, we can try to vary both n and x in fn (x). We begin with the first version of convergence. Definition 7.1. Let {fn } be a sequence of functions and f be a function defined on S ⊆ R. Then we say the sequence converges pointwise to f on S if, for each x in 157

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S, the sequence {fn (x)} converges to f (x). In this case, we say f is the pointwise limit of the sequence on S and write fn → f pointwise

or

f = lim fn . n→∞

So fn → f pointwise on S means, for each x in S and each  > 0, there exists N such that n≥N

implies ∣fn (x) − f (x)∣ < .

The key point here is that this N depends on both x and . In other words, even for a fixed  > 0, two different points x and y would normally require two different integers, say N (x) and N (y). As x varies through the domain S, the set of N (x) may well be unbounded. Thus, there may not be a single integer N that can work for all the points x in S. To see whether pointwise convergence is a good enough concept of convergence of sequences of functions, let us consider some examples. We will examine what properties, if any, are automatically transferred from the functions fn to the pointwise limit. The reader should try to fill in the details of the following examples. 7.1.2

Examples of Pointwise Convergence

The first example demonstrates that the property of having a certain limit at a point is not usually transferred to the pointwise limit. Example 7.1. Consider fn = xn on [0, 1). Then fn → f = 0 pointwise on [0, 1). The point 1 is a limit point of [0, 1). But for each n we have limx→1 fn = 1, while limx→1 f = 0. Therefore, we have lim lim fn = 0 =/ 1 = lim lim fn .

x→1 n→∞

n→∞ x→1

(7.1)

So the two limit operations, limx→1 and limn→∞ , do not commute. This is another way of saying that the property of having limit 1 as x → 1 is not transferred to the pointwise limit. To say it yet another way, preservation of the limit operation limx→c is really about the commutation of limx→c and limn→∞ . The next example shows that continuity is not usually transferred to the pointwise limit. Example 7.2. Modify the previous example, and consider fn = xn on [0, 1]. Define f on [0, 1] by f (x) = 0 if 0 ≤ x < 1 and f (1) = 1. Then fn → f pointwise on [0, 1]. Each fn is continuous at 1, but f is not because lim lim fn = lim f (x) = 0 =/ 1 = f (1) = lim lim fn .

x→1 n→∞

x→1

n→∞ x→1

Therefore, we again have the non-commutation of limit operations, this time expressing the non-preservation of continuity in the pointwise limit.

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The next example shows that integrals do not behave nicely with respect to pointwise convergence. Example 7.3. For n ≥ 2 consider the functions fn on [0, 1] defined as ⎧ ⎪ 0 if 0 ≤ x ≤ 1 − n2 , ⎪ ⎪ ⎪ ⎪ 2 fn (x) = ⎨n (x − 1) + 2n if 1 − n2 ≤ x ≤ 1 − n1 , ⎪ ⎪ ⎪ 2 ⎪ if 1 − n1 ≤ x ≤ 1. ⎪ ⎩−n (x − 1) The easiest way to understand fn is to look at its graph, which has a triangular spike over [1 − n2 , 1] with height n. Each fn is continuous, and hence integrable, on 1 [0, 1] with ∫0 fn = 1. On the other hand, we have fn → f = 0 pointwise on [0, 1] 1 and ∫0 f = 0. Therefore, we have 1

∫

0

1

lim fn = 0 =/ 1 = lim ∫ n→∞ n→∞

0

fn . 1

This non-commutation of the limit operation limn→∞ and the integral ∫0 says that the integral of the pointwise limit does not need to be the limit of the sequence of integrals. The following example illustrates that derivatives do not behave nicely with respect to pointwise convergence. Example 7.4. Consider the functions fn = n1 sin(n2 x) on [0, 1]. Then fn′ (x) = n cos(n2 x), so limn→∞ fn′ (x) does not exist for any x. On the other hand, we have fn → f = 0 pointwise on [0, 1], so f ′ = 0. Therefore, we have ′

[ lim fn (x)] = 0 =/ lim fn′ (x). n→∞

n→∞

This non-commutation of the limit operation limn→∞ and the derivative says that the derivative of the pointwise limit does not need to be the limit of the sequence of derivatives. As the examples above illustrate, pointwise convergence is not strong enough for the pointwise limit to inherit good properties from the functions fn . We now introduce a stronger form of convergence, which will be studied further in the next section, that ensures that certain properties are preserved in the limit. 7.1.3

Uniform Convergence

Definition 7.2. Let {fn } be a sequence of functions and f be a function defined on S ⊆ R. Then we say the sequence converges uniformly to f on S if, for each  > 0, there exists an integer N such that n≥N

implies ∣fn (x) − f (x)∣ < 

for all x in S.

In this case, we say f is the uniform limit of the sequence on S and write fn → f uniformly.

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It is an easy exercise to see that uniform convergence implies pointwise convergence. Therefore, when fn → f uniformly on S, we can also write f = limn→∞ fn . The key point of uniform convergence is that, given  > 0, there is one integer N that works for all x in S. Intuitively, this says that the graphs of fn for all n sufficiently large are within an -tube of the graph of f . We will show in the next section that uniform convergence forces the passage of several good properties to the uniform limit. 7.1.4

Examples of Uniform Convergence

Example 7.5. In Example 7.1, the sequence fn = xn does not convergence uniformly to f = 0 on [0, 1). Indeed, set  = 21 . Then for any integer N > 0, any point 1 x satisfying 2− N ≤ x < 1 yields 1 ∣xN − 0∣ ≥ = . 2 This shows that fn → f pointwise but not uniformly on [0, 1). Example 7.6. In Example 7.4, the sequence fn = n1 sin(n2 x) converges uniformly to f = 0 on [0, 1]. Indeed, given  > 0, simply choose N such that N  > 1. Then for n ≥ N and all x in [0, 1], we have 1 1 1 < . ∣ sin(n2 x) − 0∣ ≤ ≤ n n N Therefore, we have fn → f uniformly on [0, 1]. Example 7.4 and this example together imply that uniform convergence is still not strong enough to ensure the preservation of derivative in the uniform limit. Example 7.7. On [0, 1] consider the function fn = uniformly on [0, 1]. Moreover, since fn′ = n1 , we have

x . n

Then we have fn → f = 0

′

[ lim fn (x)] = 0 = lim fn′ (x), n→∞

n→∞

unlike Example 7.4. Therefore, in this case the limit operation limn→∞ and the derivative do commute. 7.1.5

Uniformly Cauchy

Definition 7.2 can be hard to use in practice because one has to know the uniform limit f in advance to show that fn converges to it uniformly. Analogous to Theorem 2.10, we would like to describe a uniformly convergent sequence of functions without having to know the uniform limit in advance. To achieve this we need the following concept, which is the functional counterpart of a Cauchy sequence. Definition 7.3. A sequence of functions {fn } on S is said to be uniformly Cauchy on S if, for every  > 0, there exists N such that n, m ≥ N implies ∣fn (x) − fm (x)∣ <  for all x in S.

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The following Cauchy criterion for uniform convergence is the functional counterpart of Theorem 2.10. Theorem 7.1. A sequence of functions {fn } converges uniformly to a function f on S if and only if it is uniformly Cauchy on S. Ideas of Proof. Both directions are 2 -arguments similar to the proofs of Theorems 2.9 and 2.10. Proof. that

For the “only if” part, suppose  > 0 is given. Then there exists N such n≥N

implies ∣fn (x) − f (x)∣ <

 2

for all x in S.

Thus, for n, m ≥ N and x in S, we have ∣fn (x) − fm (x)∣ ≤ ∣fn (x) − f (x)∣ + ∣f (x) − fm (x)∣ < , showing that {fn } is uniformly Cauchy. To prove the “if” part, assume that {fn } is uniformly Cauchy on S. For the function f , first note that {fn (x)} is a Cauchy sequence, and hence a convergent sequence, for each x in S (Exercise (5) below). We may therefore define f (x) = lim fn (x) n→∞

for each x in S.

To see that we have uniform convergence, suppose  > 0 is given. Then there exists N such that  for all x in S. n, m ≥ N implies ∣fn (x) − fm (x)∣ < 2 We have the inequality ∣fn (x) − f (x)∣ ≤ ∣fn (x) − fm (x)∣ + ∣fm (x) − f (x)∣, which holds for all m, n and x in S. The first term on the right is < 2 for all x as long as n, m ≥ N . Regardless of what x is, a large enough m, depending on x, will make the second term on the right also < 2 . Therefore, we have shown that fn → f uniformly on S.  We will make use of Theorem 7.1 in the next section to show that uniform convergence preserves certain functional properties. 7.1.6

Exercises

(1) Write down explicitly what it means for a sequence of functions {fn } to not converge pointwise or uniformly to a function f on S. (2) Prove that if {fn } converges uniformly to f on S, then fn converges pointwise to f on S.

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(3) Prove that {fn } converges uniformly to f on S if and only if lim (sup {∣fn (x) − f (x)∣ ∶ x ∈ S}) = 0.

n→∞

(4) Prove that the uniform limit of a uniformly convergent sequence of functions is unique. In other words, if {fn } converges uniformly to f on S and {fn } converges uniformly to g on S, then f = g on S. (5) Suppose {fn } is uniformly Cauchy on S. Prove that {fn (x)} is a Cauchy sequence for each x in S. (6) In each case, determine whether {fn } converges pointwise, uniformly, or neither. (a) (b) (c) (d) (e) (f)

fn = fn = fn = fn = fn = fn =

x n

on [0, b] or [0, ∞).

x on [0, 1] or [0, 1). 1+nx nx on [0, 1] or [0, ∞). 1+nx nx on [0, b] or [0, ∞). 1+n3 x x on [0, ∞). x+n nx on [0, b] or [0, ∞). nx e that in Example 7.1 fn = xn

(7) Prove → f = 0 pointwise on [0, 1). 1 (8) Prove that in Example 7.2 limn→∞ fn = ∫0 f . (9) In the context of Example 7.3: (a) Sketch the graph of fn . Then prove that it is continuous on [0, 1]. 1 (b) Prove that ∫0 fn = 1. (c) Prove that fn → f = 0 pointwise on [0, 1]. (10) In the context of Example 7.4: (a) (b) (c) (d)

Sketch the graph of fn . Compare the graphs of f1 , f2 , and f3 . Prove that limn→∞ fn′ (x) does not exist for any x. Prove that fn → f = 0 pointwise on [0, 1]. 1 1 Prove that limn→∞ ∫0 fn = ∫0 f .

(11) In the context of Example 7.7: (a) Sketch the graph of fn . Compare the graphs of f1 , f2 , and f3 . (b) Prove that fn → f = 0 uniformly on [0, 1]. 1 1 (c) Prove that limn→∞ ∫0 fn = ∫0 f . (12) Suppose each fn is an increasing function and that {fn } converges pointwise to f on [a, b]. Prove that f is increasing. (13) Suppose {fn } converges uniformly to f on S with each fn continuous and that {an } is a sequence in S converging to a in S. Prove that the sequence {fn (an )} converges to f (a). (14) In the previous exercise, give an example to show that {fn (an )} may not converge to f (a) if uniform convergence is replaced by pointwise convergence. (15) Prove that a sequence {fn } is uniformly Cauchy if and only if, for every  > 0, there exists N such that n>m≥N

implies ∣fn (x) − fm (x)∣ < 

for all

x ∈ S.

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In this section we discuss several functional properties that are preserved by uniform convergence. As explained in the previous section, each such preservation result can be interpreted as a commutation property between the limit operation limn→∞ and the functional property under consideration. Another main theme in this section is that we will recycle essentially the same 3 -argument many times. 7.2.1

Preservation of Limits

Example 7.1 shows that pointwise convergence is not strong enough to force the limit operation limx→c to pass from fn to the pointwise limit. The following result shows that uniform convergence is strong enough for this purpose. Theorem 7.2. Suppose fn converges uniformly to f on S, c is a limit point of S, and that the limit an = limx→c fn (x) exists for each n. Then the sequence {an } is convergent and lim f (x) = lim an .

x→c

n→∞

Ideas of Proof. Both assertions are 3 -arguments. To see that {an } is Cauchy, we use the estimates an ≈ fn (x) ≈ fm (x) ≈ am . Here each wavy equal sign ≈ means that the two quantities next to it can be closely approximated. The first and the last ≈ hold if x is sufficiently close to c, while the middle ≈ holds for large enough n and m. For the equality of limits, we use the estimates lim an ≈ an ≈ fn (x) ≈ f (x).

n→∞

The first and the last ≈ hold for large enough n, while the middle ≈ holds when x is close enough to c. Proof. To prove that {an } is convergent, it suffices to show that it is a Cauchy sequence. Given  > 0, the uniform convergence assumption and Theorem 7.1 imply that there exists N such that  for all x in S. n, m ≥ N implies ∣fn (x) − fm (x)∣ < 3 For any such n and m, we also know that there exist δ1 and δ2 > 0 such that  0 < ∣x − c∣ < δ1 implies ∣fn (x) − an ∣ < 3 and  0 < ∣x − c∣ < δ2 implies ∣fm (x) − am ∣ < , 3

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provided x is in S. Thus, as long as n, m ≥ N and 0 < ∣x − c∣ < min{δ1 , δ2 } with x in S, the inequality ∣an − am ∣ ≤ ∣an − fn (x)∣ + ∣fn (x) − fm (x)∣ + ∣fm (x) − am ∣ implies ∣an − am ∣ < . Therefore, {an } is a Cauchy sequence, and hence a convergent sequence. The assertion about the equality of limits is proved by a similar 3 -argument, which is left as an exercise.  The equality of limits in Theorem 7.2 can be written as lim lim fn (x) = lim lim fn (x).

x→c n→∞

n→∞ x→c

In other words, under the hypothesis of uniform convergence, the limit operations limn→∞ and limx→c commute. 7.2.2

Preservation of Integrals

Example 7.3 shows that pointwise convergence is not strong enough to force the integral to commute with the limit operation limn→∞ . The following result shows that uniform convergence is strong enough for this purpose. Theorem 7.3. Suppose fn converges uniformly to f on [a, b], and each fn is integrable on [a, b]. Then f is also integrable on [a, b], and b

∫

a

b

f = lim ∫ n→∞

a

fn .

Ideas of Proof. The integrability of f is again proved by an the approximations

 -argument, 3

using

U (f ; P ) ≈ U (fn ; P ) ≈ L(fn ; P ) ≈ L(f ; P ). The two outer ≈ hold for large enough n, while the middle ≈ holds for partitions P with small enough norms. The equality is proved using the approximation f ≈ fn for large enough n and Corollary 6.3. Proof. To prove that f is integrable on [a, b], suppose  > 0 is given. By uniform convergence there exists N such that  for all x in [a, b]. n ≥ N implies ∣fn (x) − f (x)∣ < 3(b − a) Fix such an n, for example, n = N . The integrability of fn implies the existence of a partition P of [a, b] such that  U (fn ; P ) − L(fn ; P ) < . 3

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In each sub-interval of P , we have  ∣ui (f ) − ui (fn )∣ ≤ 3(b − a)

and ∣li (f ) − li (fn )∣ ≤

165

 . 3(b − a)

Therefore, the inequalities U (f ; P ) − L(f ; P ) ≤ ∣U (f ; P ) − U (fn ; P )∣ + ∣U (fn ; P ) − L(fn ; P )∣ + ∣L(fn ; P ) − L(f ; P )∣  < ∑ ∣ui (f ) − ui (fn )∣∆xi + + ∑ ∣li (fn ) − li (f )∣∆xi 3    ≤ (b − a) + + (b − a) =  3(b − a) 3 3(b − a) imply that f is integrable on [a, b]. For the equality, suppose  > 0 is given. Then uniform convergence implies that there exists N such that  for all x in [a, b]. n ≥ N implies ∣fn (x) − f (x)∣ < b−a For such n, the inequalities   + fn (x) < f (x) < + fn (x) − b−a b−a and Corollary 6.3 imply b

b

fn ≤ ∫

− + ∫

a

b

f ≤+∫

a

a

fn .

This is equivalent to b

b

∣∫

a

f −∫

a

fn ∣ ≤ .

The desired equality follows.



The equality in Theorem 7.3 can be written as b

∫

a

b

lim fn = lim ∫ n→∞ n→∞

a

fn . b

In other words, under the hypothesis of uniform convergence, the integral ∫a and the limit operation limn→∞ commute. 7.2.3

Preservation of Continuity

Example 7.2 shows that pointwise convergence is not strong enough to preserve continuity. The following result shows that uniform convergence is strong enough for this purpose. Theorem 7.4. Suppose fn converges uniformly to f on S, and each fn is continuous at a point c in S. Then f is also continuous at c.

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Ideas of Proof. It is possible to obtain this result using Theorem 7.2; the details are left as an exercise. Here we provide a direct proof using an 3 -argument via the approximations f (x) ≈ fn (x) ≈ fn (c) ≈ f (c). The two outer ≈ hold for large enough n. The middle ≈ holds for x close to c. Proof.

Given  > 0, uniform convergence implies there exists N such that  for all x in S. n ≥ N implies ∣f (x) − fn (x)∣ < 3 Fix such an n, for example, n = N . Continuity of fn at c implies there exists δ > 0 such that  ∣x − c∣ < δ implies ∣fn (x) − fn (c)∣ < , 3 provided x is in S. Therefore, the inequalities ∣f (x) − f (c)∣ ≤ ∣f (x) − fn (x)∣ + ∣fn (x) − fn (c)∣ + ∣fn (c) − f (c)∣ <  hold whenever ∣x − c∣ < δ with x in S. This shows that f is continuous at c.



When c is a limit point of S, using (4.1) the conclusion of the previous theorem can be written as lim lim fn (x) = lim f (x) = f (c) = lim fn (c) = lim lim fn (x).

x→c n→∞

x→c

n→∞

n→∞ x→c

In other words, under the hypothesis of uniform convergence, the limit operations limn→∞ and limx→c commute. 7.2.4

Preservation of Derivatives

Examples 7.4 and 7.6 together show that uniform convergence is not strong enough for the preservation of derivatives. One way to ensure derivatives are preserved is to have uniform convergence for the derivatives and convergence at a single point. Theorem 7.5. Suppose each fn is differentiable on an open interval I containing [a, b], there exists a point α in [a, b] such that {fn (α)} is a convergent sequence, and fn′ converges uniformly to a function g on [a, b]. Then: (1) fn converges uniformly to a function f on [a, b]. (2) f is differentiable on (a, b) with f ′ = g. Ideas of Proof. The Mean Value Theorem 5.7 is a way to relate a function and its derivative, so it is no surprise that we will use it. The first assertion is proved by an 2 -argument. The other assertion is proved by an 3 -argument via the approximations f (x) − f (d) fm (x) − fm (d) ′ ≈ ≈ fm (d) ≈ g(d). x−d x−d

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The last ≈ holds for large enough m, while the middle ≈ holds for x close enough to d. The first ≈ requires a bit of work. Proof. For the first assertion, we will show that {fn } is uniformly Cauchy, which is enough by Theorem 7.1. For each point x in [a, b] and any two indices n and m, the Mean Value Theorem applied to fn − fm yields a point c between x and α such that ′ fn (x) − fm (x) = [fn (α) − fm (α)] + (x − α)[fn′ (c) − fm (c)].

We know that {fn (α)} is a Cauchy sequence, and {fn′ } is uniformly Cauchy. Thus, given  > 0, there exist N1 and N2 such that  n, m ≥ N1 implies ∣fn (α) − fm (α)∣ < 2 and  ′ (y)∣ < n, m ≥ N2 implies ∣fn′ (y) − fm for all y in S. 2(b − a) Since ∣x − α∣ ≤ b − a, for n, m ≥ max{N1 , N2 }, we have ′ ∣fn (x) − fm (x)∣ ≤ ∣fn (α) − fm (α)∣ + ∣x − α∣∣fn′ (c) − fm (c)∣   = . < + (b − a) 2 2(b − a)

This proves that {fn } converges uniformly to a function f on S. Pick a point d in (a, b), and we will show f ′ (d) = g(d). For any two indices n and m, the Mean Value Theorem applied to fn − fm on the interval determined by d and x =/ d in (a, b) yields a point e between d and x such that fn (x) − fn (d) fm (x) − fm (d) ′ − = fn′ (e) − fm (e). x−d x−d Given  > 0, using the above equality, the uniform convergence of {fk′ } now implies there exists N1 such that n, m ≥ N1

implies

∣

fn (x) − fn (d) fm (x) − fm (d)  − ∣< . x−d x−d 3

At this moment d and x are fixed. So keeping m ≥ N1 fixed and taking the limit limn→∞ in the above inequality, we obtain ∣

f (x) − f (d) fm (x) − fm (d)  − ∣≤ . x−d x−d 3

′ Since g(d) = limm→∞ fm (d), there exists N2 such that

 ′ ∣fm (d) − g(d)∣ < . 3 Now set m = max{N1 , N2 }, so the previous two inequalities both hold. The differentiability of fm at d implies the existence of δ > 0 such that m ≥ N2

0 < ∣x − d∣ < δ

implies

implies

∣

 fm (x) − fm (d) ′ − fm (d)∣ < , x−d 3

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provided x is in (a, b). Therefore, for such x, the last three 3 -inequalities together imply ∣

f (x) − f (d) − g(d)∣ < . x−d

This proves the differentiability of f at d and f ′ (d) = g(d).



The assertion f ′ = g in the previous theorem can be written as dfn d ( lim fn ) = lim . n→∞ dx dx n→∞ In other words, under the stated hypotheses, the differential operator limit limn→∞ commute. 7.2.5

d dx

and the

Exercises

(1) If {fn } and {gn } converge uniformly on S to f and g, respectively, prove that {fn + gn } converges uniformly on S to f + g. (2) In the previous exercise, give an example to show that {fn gn } may not converge uniformly on S to f g. (3) Suppose {fn } and {gn } converge uniformly on S to f and g, respectively, and that all the fn and gn are bounded on S. Prove that {fn gn } converges uniformly on S to f g. (4) Give an example with the following properties. The sequence {fn } converges to f pointwise on an interval [a, b], each fn is integrable on [a, b], and f is not integrable on [a, b]. (5) Suppose fn converges uniformly to f on S, and each fn is uniformly continuous on S. Prove that f is uniformly continuous on S. (6) Prove the equality of limits in Theorem 7.2. Also, where exactly in that proof is the hypothesis that c is a limit point of S used? (7) Prove the following statement that was used in the proof of Theorem 7.3. If ∣g(x) − f (x)∣ < δ

for all

x ∈ S,

then ∣sup{g(x) ∶ x ∈ S} − sup{f (x) ∶ x ∈ S}∣ ≤ δ and ∣inf{g(x) ∶ x ∈ S} − inf{f (x) ∶ x ∈ S}∣ ≤ δ. (8) The proof of Theorem 7.4 uses the -δ characterization of continuity. Write down a version of that proof that uses Definition 4.5 of continuity. (9) Use Theorem 7.2 to give another proof of Theorem 7.4. (10) Give an example to show that the conclusion in Theorem 7.2 may still hold even if fn does not converge uniformly to f .

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(11) Give an example to show that the conclusion in Theorem 7.3 may still hold even if fn does not converge uniformly to f . (12) Give an example in which fn is not continuous at any point on [0, 1], and fn converges uniformly to a continuous function on [0, 1]. (13) Suppose {fn } converges uniformly to f on S and that each fn is bounded on S. Prove that f is bounded on S. (14) In the previous exercise, give an example to show that f may not be bounded if uniform convergence is replaced by pointwise convergence. (15) A sequence of functions {fn } on S is called uniformly bounded if there exists a real number M such that ∣fn (x)∣ ≤ M for all n and x in S. Suppose {fn } converges uniformly to f on S and that each fn is bounded on S. Prove that {fn } is uniformly bounded. (16) In the previous exercise, give an example to show that {fn } may not be uniformly bounded if uniform convergence is replaced by pointwise convergence.

7.3

Series of Functions

As discussed in chapter 2.5, a series is a particular kind of a sequence in which the nth term is a partial sum of n quantities. In this section we consider series of functions. We first establish a Cauchy Criterion, the Weierstrass M -Test, and some results about interchange of limits for series of functions. Then we construct a function that is continuous on R but is nowhere differentiable. 7.3.1

Definition

Definition 7.4. Suppose {fn } is a sequence of functions on S. The sequence of functions {sn } on S with each m

sm = f1 + ⋯ + fm = ∑ fi i=1

is called a series of functions, also written as ∑ fn . The function sm is called the mth partial sum. If {sn } converges pointwise or uniformly to a function f on S, then we also write f = ∑ fn . Convergence results about series of real numbers and sequences of functions can now be used to obtain convergence results for series of functions. 7.3.2

Cauchy Criterion

Applying Theorem 7.1 to a series of functions, we obtain the following Cauchy criterion for uniform convergence.

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Corollary 7.1. A series of functions ∑ fn converges uniformly to a function on S if and only if, for every  > 0, there exists N such that n

n>m≥N

implies

∣ ∑ fi (x)∣ < 

for all x in S.

i=m+1

Proof.

First observe that n

sn − sm = ∑ fi i=m+1

for n > m. Therefore, the stated condition says exactly that the sequence of functions {sn } is uniformly Cauchy, which by Theorem 7.1 is equivalent to uniform convergence.  A particularly useful consequence of the previous Cauchy criterion is the following test for uniform convergence. 7.3.3

Weierstrass M -Test

Corollary 7.2. Suppose for each n there exists a real number Mn such that ∣fn (x)∣ ≤ Mn

for all x in S.

If the series ∑ Mn converges, then the series of functions ∑ fn converges uniformly on S. Proof. implies

By Theorem 3.1, for every  > 0, there exists N such that n > m ≥ N n

n

n

∣ ∑ fi (x)∣ ≤ ∑ ∣fi (x)∣ ≤ ∑ Mi <  i=m+1

i=m+1

i=m+1

for all x in S. Corollary 7.1 now says that the series of functions ∑ fn converges uniformly.  Example 7.8. By the Weierstrass M -Test, the series ∑

1 (x + n)p

converges uniformly on [0, ∞) if p > 1, since ∑ n1p is a convergent p-series. Be careful that, unlike the Cauchy criterion, the Weierstrass M -Test is a oneway implication. So if the series ∑ Mn diverges, then there is no conclusion to be made from the Weierstrass M -Test.

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Interchange of Limits

Results about interchange of limits in the previous section can be applied to series of functions. The proofs of the next few results are immediate applications of the corresponding results in the previous section. The following result is the special case of Theorem 7.2 when applied to a series of functions. Corollary 7.3. Suppose a series of functions ∑ fn converges uniformly to f on S, c is a limit point of S, and that the limit an = limx→c fn (x) exists for each n. Then the series ∑ an is convergent and lim f (x) = ∑ an .

x→c

The equality of the previous corollary may be written as lim (∑ fn (x)) = ∑ (lim fn (x)) .

x→c

x→c

In other words, under the hypothesis of uniform convergence, the limit operation limx→c and the series operation ∑ commute. The next result is the special case of Theorem 7.3 when applied to a series of functions. Corollary 7.4. Suppose a series of functions ∑ fn converges uniformly to f on [a, b], and each fn is integrable on [a, b]. Then f is also integrable on [a, b], the b series ∑ (∫a fn ) is convergent, and b

∫

a

b

f = ∑ (∫

a

fn ) .

The equality of the previous corollary may be written as b

∫

a

b

(∑ fn ) = ∑ (∫

a

fn ) . b

In other words, under the hypothesis of uniform convergence, the integral ∫a and the series operation ∑ commute. This is saying that the integral of the limit can be computed by first integrating each term. The following result is the special case of Theorem 7.4 when applied to a series of functions. Corollary 7.5. Suppose a series of functions ∑ fn converges uniformly to f on S, and each fn is continuous at a point c in S. Then f is also continuous at c. When c is a limit point of S, the conclusion of the previous theorem can be written as lim (∑ fn (x)) = ∑ (lim fn (x)) .

x→c

x→c

The next result is the special case of Theorem 7.5 when applied to a series of functions.

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Corollary 7.6. Suppose each fn is differentiable on an open interval I containing [a, b], there exists a point α in [a, b] such that ∑ fn (α) is a convergent series, and ∑ fn′ converges uniformly to a function g on [a, b]. Then: (1) ∑ fn converges uniformly to a function f on [a, b]. (2) f is differentiable on (a, b) with f ′ = ∑ fn′ . The equality in the previous corollary may be written as dfn d (∑ fn ) = ∑ ( ). dx dx d In other words, under the stated hypotheses, the differential operator dx commutes with the series operation ∑. This is saying that the derivative of the limit can be computed by differentiating each term fn .

7.3.5

A Continuous Nowhere Differentiable Functions

Here we construct a continuous function on R that is not differentiable at any point. There are two purposes of this example. First, it shows that some weird things can happen for a continuous function. In fact, it almost defies the imagination that a continuous function on R can be nowhere differentiable. Second, the construction is a good illustration of the Weierstrass M -Test and Corollary 7.5. Let us first explain the ideas for the construction. We will construct a continuous function f = ∑ fn . With carefully chosen values hn that converge to 0, the difference quotients h1n (f (x + hn ) − f (x)) can be made to jump between odd and even integers as n increases. This suffices to show that f is not differentiable at x. For the actual construction, we need the following definition. Definition 7.5. Let p be a positive real number. A function f on R is called p-periodic if it satisfies f (x + p) = f (x) for all x in R. A p-periodic function is uniquely determined by its values in the interval [0, p]. Conversely, given a function f on [0, p] with f (0) = f (p), there exists a unique p-periodic function on R that agrees with f on [0, p]. Now let f0 be the 2-periodic function uniquely determined by ⎧ ⎪ ⎪x f0 (x) = ⎨ ⎪ ⎪ ⎩2 − x

if 0 ≤ x ≤ 1, if 1 ≤ x ≤ 2.

For k ≥ 1 define fk (x) =

1 f0 (4k x), 4k

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which is 42k -periodic. Graphically, f0 consists of the [−1, 1] portion of the graph of ∣x∣, which is then made 2-periodic. Over [0, 2] its graph has the shape of a pyramid of height 1. It consists of two lines, one with slope 1 and the other with slope −1. Over [0, 21 ] the graph of f1 is a pyramid of height 14 , and four of these are inscribed inside the pyramid of f0 over [0, 2]. The lines in its graph have slopes either 1 or −1. In general, the graph of fk+1 consists of little pyramids that are four times as small, both in width and in height, as those in fk . Each function fk is continuous on R and satisfies 1 for all x in R. 4k Therefore, the Weierstrass M -Test (Corollary 7.2) says the series of functions ∑∞ i=0 fi converges uniformly to a function f on R. Moreover, by Corollary 7.5 the function f is continuous on R. Now we observe that f is not differentiable at any point. Pick a point c in R, and we will show that f is not differentiable at c. For each integer k ≥ 0, define ∣fk (x)∣ ≤

hk = ±

2 , 4k+1

where the sign ± is chosen such that both 4k c and 4k (c + hk ) lie in the interval [m, m + 1] for some integer m. This is possible because 4k hk = ± 12 . For each k, the choice of hk and the 42i -periodicity of fi imply ⎧ ⎪ ⎪±hk fi (c + hk ) − fi (c) = ⎨ ⎪ ⎪ ⎩0

if 0 ≤ i ≤ k, if i > k.

(7.2)

For example, we have fk+1 (c + hk ) − fk+1 (c) = 0 because fk+1 is

2 -periodic. 4k+1

Likewise, we have

f0 (4k (c + hk )) − f0 (4k c) = ±hk . 4k Therefore, the relevant difference quotient of f is fk (c + hk ) − fk (ck ) =

f (c + hk ) − f (c) k fi (c + hk ) − fi (c) k =∑ = ∑ ±1. hk hk i=0 i=0

(7.3)

This last quantity is odd when k is even, and it is even when k is odd. Therefore, this sequence of difference quotients does not converge. Since the sequence {hk } (c) converges to 0, we conclude that limh→0 f (c+h)−f does not exist. h 7.3.6

Exercises

(1) Determine whether the following series converge uniformly. 2

(a) ∑ cosnn2

x

on R.

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(b) (c) (d) (e) (f) (2) (3) (4) (5) (6) (7) (8)

(9) (10)

(11) (12) (13)

∑ x1n on [1, 2] or [2, ∞). ∑ xnn on [1, 2] or [2, ∞). ∑ xnn on [−r, r] for 0 < r < 1 or (−1, 1). ∑ xn2 on [−1, 1] or R. n ∑ ( x2 ) on [−r, r] for 0 < r < 2 or (−2, 2).

Write down the details of the proof of Corollary 7.3. Write down the details of the proof of Corollary 7.4. Write down the details of the proof of Corollary 7.5. Write down the details of the proof of Corollary 7.6. Suppose ∑ fn converges uniformly to f on S, and each fn is uniformly continuous on S. Prove that f is also uniformly continuous on S. Suppose ∑ fn and ∑ gn converge uniformly to f and g on S, respectively. Prove that ∑(fn + gn ) converges uniformly to f + g on S. Suppose ∑ fn converges uniformly to f on S. Prove that the sequence of functions {fn } converges uniformly to the 0 function. This is the functional analog of Corollary 3.2. Prove that a p-periodic function f is uniquely determined by its values in [0, p], and f (0) = f (p). Suppose f is a function on [0, p] such that f (0) = f (p). Prove that there exists a unique p-periodic function g on R such that g(x) = f (x) for x in [0, p]. The function g is called the periodic extension of f . Suppose f is a p-periodic function, and k is a positive integer. Prove that f is kp-periodic. Suppose f is a p-periodic function, and r is a positive real number. Prove that g(x) = f (rx) is pr -periodic. In section 7.3.5: (a) Sketch the graphs of f0 , f1 , and f2 . (b) Prove that each fk is continuous on R. (c) For each point c in R and integer k ≥ 0, prove that there exists a sign, + or 2 −, such that both 4k c and 4k (c ± 4k+1 ) lie in the interval [m, m + 1] for some integer m. (d) Justify (7.2) and (7.3).

(14) This exercise provides another continuous nowhere differentiable function using a variation of the construction in section 7.3.5. With the same notations there, define ∞ 3 k g(x) = ∑ ( ) f0 (4k x). k=0 4 (a) Prove that g is continuous on R. (b) Prove that ∣

k−1 g(c + hk ) − g(c) 3k + 1 ∣ ≥ 3k − ∑ 3i = . hk 2 i=0

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(c) Use the previous part to conclude that g is not differentiable at c.

7.4

Power Series

In this section, we discuss power series. They are particular examples of series of functions that have very nice properties. We will study two issues. First, given a power series, find where it is pointwise or uniformly convergent, and see whether its derivative and integral can be computed term-wise. Second, given a function, find where it can be represented by a power series. Let us begin with some definitions.

7.4.1

Definitions

Definition 7.6. Given a point c in R, a power series at c is a series of functions of the form ∞ i ∑ ai (x − c) = a0 + a1 (x − c) + ⋯, i=0

where each ai is a real number. We will often omit “at c” when we talk about power series. Observe that a power series is a series of functions ∑ fi in which fi = ai (x − c)i is a degree i polynomial, provided ai =/ 0. In particular, each fi is defined on R and has all higher derivatives. Moreover, a power series at c must converge at the point x = c, since fi (c) = 0 for all i > 0. The main question is where a power series converges besides the point x = c. First recall from section 3.4 the concept of absolute convergence, the Root Test, and the Ratio Test. We will also use the concept of limit superior discussed in section 2.4. Let us briefly motivate the next definition. Since fi = ai (x − c)i is a degree i polynomial, this suggests a comparison with a geometric series ∑ ri , so we have the estimate ai (x − c)i ≈ ri . The geometric series is convergent exactly when ∣r∣ < 1. Thus, taking the ith root above suggests that a condition similar to 1 for large i ∣x − c∣ < 1 ∣ai ∣ i should guarantee the absolute convergence of the power series. This leads to the following definitions. 1

Definition 7.7. For a power series ∑ ai (x − c)i , write L = lim sup ∣an ∣ n . Define its radius of convergence as the extended real number ⎧ ⎪ 0 if L = ∞, ⎪ ⎪ ⎪ ⎪1 R = ⎨L if 0 < L < ∞, ⎪ ⎪ ⎪ ⎪ ⎪ ⎩+∞ if L = 0.

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Its interval of convergence is defined as the open interval (c − R, c + R) if R > 0, and as the single point {c} if R = 0. 7.4.2

Convergence

The discussion just before the previous definition is made precise in the following convergence theorem. Theorem 7.6. Suppose ∑ ai (x − c)i is a power series with radius of convergence R. (1) The power series is absolutely convergent for x in the interval of convergence. (2) The power series is divergent if ∣x − c∣ > R. Proof.

We will prove the case 1

0 < L = lim sup ∣an ∣ n < ∞. The cases L = 0 and L = ∞ are exercises. By the Root Test (Theorem 3.7), the power series is absolutely convergent if 1

1

lim sup ∣an (x − c)n ∣ n = ∣x − c∣ (lim sup ∣an ∣ n ) = ∣x − c∣L < 1. This is equivalent to saying x is in the interval of convergence. The Root Test also says that the power series is divergent if 1

lim sup ∣an (x − c)n ∣ n = ∣x − c∣L > 1. This proves the theorem.



As the reader learned in calculus, the convergence behavior of a power series at the end points c±R of the interval of convergence needs to be determined separately. The convergence tests developed in chapter 2.5 can be used for this purpose. One must be careful that Theorem 7.6 only tells us where the power series converges pointwise, namely, the interval of convergence, possibly including the end points. It does not say where it converges uniformly, which we will discuss shortly. To determine the radius and interval of convergence, one has to compute 1 lim sup ∣an ∣ n , which may not be an easy task. In practice one can often get away with an easier computation as follows. As a consequence of Theorem 2.14 and Exercise (6) on page 76, if the limit an+1 L′ = lim ∣ ∣ (7.4) n→∞ an 1 exists as an extended real number, then it is equal to lim ∣an ∣ n . Therefore, we have the following result about the radius of convergence. Theorem 7.7. Suppose ∑ ai (x − c)i is a power series such that the limit L′ in (7.4) exists as an extended real number. Then the radius of convergence is equal to ⎧ ⎪ 0 if L′ = ∞, ⎪ ⎪ ⎪ ⎪1 ⎨ L′ if 0 < L′ < ∞, ⎪ ⎪ ⎪ ′ ⎪ ⎪ ⎩+∞ if L = 0.

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Uniform Convergence

In section 7.3.4 we discussed several properties that are preserved by uniform convergence. The following observation will allow us to use those results. Theorem 7.8. A power series ∑ ai (x − c)i converges uniformly on every closed bounded interval inside its interval of convergence. Ideas of Proof. We will compare the power series to a convergent geometric series and then use the Weierstrass M -Test to conclude its uniform convergence. Proof. Once again we leave the R = 0 and R = +∞ cases as exercises. We will prove the case 0 < R < +∞. A closed bounded interval inside (c − R, c + R) must be contained in a closed bounded interval I = [c − R, c + R] for some 0 <  < 1. Pick any r with  < r < 1. For x in I, we have r ∣x − c∣ < rR = 1 , lim sup ∣an ∣ n which implies 1 lim sup (∣an ∣ n ∣x − c∣) < r. Therefore, there exists N such that, for x in I, we have ∣an (x − c)n ∣ < rn for all n ≥ N . Since ∣r∣ < 1, the geometric series ∑ rn is convergent. The Weierstrass M -Test (Corollary 7.2) now implies that the power series converges uniform on I.  In particular, when the radius of convergence is +∞, the power series converges uniformly on every closed bounded interval in R. Next we turn to properties of power series that can be computed term-wise.

7.4.4

Term-Wise Integration

The following result says that a power series can be integrated term-by-term within its interval of convergence. Theorem 7.9. Suppose a power series f (x) = ∑ ai (x−c)i has radius of convergence R > 0. Then ∞ x ai (x − c)i+1 for all ∣x∣ < R, f (t)dt = ∑ ∫ c i=0 i + 1 and this power series also has radius of convergence R. Proof. The equality follows from Theorem 6.9, Corollary 7.4, and Theorem 7.8, applied to the closed bounded interval with end points c and x. The assertion concerning the radius of convergence follows from the equality 1 1 an−1 n lim sup ∣an ∣ n = lim sup ∣ ∣ , n 1 which in turn is true because lim n n = 1 (Exercise (13) on page 44). 

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Term-Wise Differentiation

The following result says that a power series can be differentiated term-by-term within its interval of convergence. Theorem 7.10. Suppose a power series f (x) = ∑ ai (x − c)i has radius of convergence R > 0. Then: (1) The power series f is differentiable in its interval of convergence. (2) Its derivative is given by ∞

f ′ (x) = ∑ iai (x − c)i−1 i=1

for x in the interval of convergence (c − R, c + R). (3) The power series ∑ iai (x − c)i−1 also has radius of convergence R. Proof. We apply Theorem 7.8 and Corollary 7.6 with fn = an (x−c)n . To see that we can actually use Corollary 7.6, it suffices to show that the term-wise differentiated power series ∑ iai (x − c)i−1 has radius of convergence R. To do this, it is enough to show 1 1 lim sup ∣an ∣ n = lim sup ∣(n + 1)an+1 ∣ n . 1 This equality holds because lim(n + 1) n = 1.  Since the term-wise differentiated power series has the same radius of convergence, the same process can be applied to it to obtain the second derivative, and so forth. Corollary 7.7. Suppose a power series f (x) = ∑ ai (x−c)i has radius of convergence R > 0. Then: (1) The nth derivative f (n) exists for each n ≥ 1 in its interval of convergence. (2) Its nth derivative is given by ∞ i! f (n) (x) = ∑ ai (x − c)i−n i=n (i − n)! for x in the interval of convergence (c − R, c + R). (3) The power series in the previous part also has radius of convergence R. Proof. The case n = 1 is Theorem 7.10. The induction step also follows from Theorem 7.10.  Therefore, a power series is infinitely differentiable in its interval of convergence, and the derivatives are all computed term-wise. Recall that a power series at c must converge at x = c. The following result was informally suggested in section 5.3.1. Corollary 7.8. Suppose a power series f (x) = ∑ ai (x−c)i has radius of convergence R > 0. Then f (n) (c) an = n! for each n ≥ 0.

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Proof. Evaluate the power series for f (n) at x = c. Every term is 0, except the first one, which is f (n) (c) =

n! an . 0!

Since 0! = 1, we obtain the desired equality. 7.4.6



Taylor Series

We now turn to the question of representing a given function by a power series. The following definition is suggested by Corollary 7.8 and also by the Taylor polynomial (5.5). Definition 7.8. Suppose f is infinitely differentiable on an open interval I that contains a point c. The Taylor series of f at c is defined as the power series f (k) (c) (x − c)k k! k=0 ∞

T (f ; c)(x) = ∑ for x in I.

There is no general claim that the Taylor series converges, or that it converges to the intended function f . An optimistic guess would be that the Taylor series converges to f in its interval of convergence. However, this is not true in general. In fact, it is possible for the Taylor series to converge to a function different from f . In other words, convergence of the Taylor series itself is not enough to imply that the limit is f . One such example is in the exercises. Note that the degree n Taylor polynomial Tn (x; c) (5.5) is the nth partial sum of the Taylor series of f . Thus, the convergence of the Taylor series is really about the convergence of the sequence of Taylor polynomials. Recall from (5.4) that the nth error term is defined as the difference Rn (x; c) = f (x) − Tn (x; c). Therefore, the Taylor series converges to f precisely when the sequence of error terms converges to 0. The following result gives a convenient sufficient condition for the Taylor series to converge to the desired function. Theorem 7.11. Suppose f is infinitely differentiable on an open interval I that contains a point c. Suppose for each x in I, there exists a real number Mx such that ∣f (n) (y)∣ ≤ Mx

for all n and all y between c and x.

Then the Taylor series T (f ; c)(x) converges pointwise to f (x) on I. Proof.

For each x in I, by Taylor’s Theorem 5.8 the nth error term is Rn (x; c) =

f (n+1) (b) (x − c)n+1 (n + 1)!

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for some point b between c and x. The assumption then implies ∣Rn (x; c)∣ ≤

Mx ∣x − c∣n+1 . (n + 1)!

This last sequence converges to 0 for each x, so limn→∞ Rn (x; c) = 0. This means that the Taylor series converges to f .  For example, the previous theorem can be applied to the familiar functions ex , sin x, and cos x to show that each of their Taylor series converges to the function itself on R. 7.4.7

Exercises

(1) Prove the cases L = 0 and L = ∞ of Theorem 7.6. (2) Prove the cases R = 0 and R = +∞ of Theorem 7.8. (3) Explain why the proof of Theorem 7.8 cannot be used to show that the power series converges uniformly on its interval of convergence. (4) Write down all the details of the proof of Theorem 7.9. (5) Write down all the details of the proof of Theorem 7.10. (6) Write down all the details of the proof of Corollary 7.7. (7) Suppose the power series ∑ ai (x−c)i and ∑ bi (x−c)i both converge to a function f on some open interval containing c. Prove that an = bn for all n. This exercise shows that the coefficients an are uniquely determined by the power series in its interval of convergence. (8) Consider the function ⎧ − 1 ⎪ ⎪e x2 f (x) = ⎨ ⎪ ⎪ ⎩0

if x =/ 0, if x = 0.

(a) Prove that f is infinitely differentiable on R. (b) Prove that the Taylor series T (f ; 0) of f at 0 is the 0 function. Conclude that the Taylor series does not converge to f on any open interval containing 0. (9) Derive the Taylor series at 0 of the functions ex , sin x, and cos x. Then show that in each case the Taylor series converges to the function itself on R. 1 (10) For each integer n ≥ 1, derive the Taylor series at 0 of (1−x) n and determine its interval of convergence. Then prove that the Taylor series converges to the function itself on the interval of convergence. (11) Repeat the previous exercise for the functions ln(1 + x) and tan−1 x. (12) Suppose f (x) = ∑ ai xi has radius of convergence R > 0. (a) Prove that f is an odd function if and only if ai = 0 for all even i. (b) Prove that f is an even function if and only if ai = 0 for all odd i.

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Additional Exercises

(1) Suppose {fn } converges uniformly to f on S and on T . Prove that {fn } converges uniformly to f on S ∪ T . (2) Suppose ∑ fn converges uniformly to f on S and on T . Prove that ∑ fn converges uniformly to f on S ∪ T . (3) Suppose ∑ fn converges uniformly to f on S with each fn bounded. Prove that f is bounded on S. (4) Suppose {fn } converges uniformly to f on S, and each fn is bounded on S. Define gn = n1 ∑ni=1 fi . Prove that {gn } converges uniformly to f on S. (5) Suppose {fn } converges to f uniformly on [a, b], each fn is integrable on [a, b], and {xn } is an increasing sequence in [a, b] that converges to b. Prove that xn

lim ∫ n→∞

a

b

fn = ∫

f.

a

(6) Give an example in which fn is not continuous at any point on [0, 1], and ∑ fn converges uniformly to a continuous function on [0, 1] (7) Suppose ∑ fn converges uniformly on S to a function f . Prove that lim (sup{∣fn (x)∣ ∶ x ∈ S}) = 0.

n→∞

(8) A sequence of functions {fn } on S is equicontinuous if, for every  > 0, there exists δ > 0 such that ∣x − y∣ < δ with x, y ∈ S

implies ∣fn (x) − fn (y)∣ < 

for all

n.

(a) Suppose {fn } is an equicontinuous sequence of functions on S that converges pointwise to f . Prove that f is uniformly continuous on S. (b) Suppose {fn } is a sequence of continuous functions on [a, b] that converges uniformly to f . Prove that {fn } is equicontinuous. (9) Abel’s Theorem says: Suppose f (x) = ∑ ai xi has radius of convergence R = 1 and that it converges at x = 1. Then it converges uniformly on [0, 1]. Prove this theorem as follows. (a) For non-negative integers m ≤ n, use the abbreviation am,n = am + am+1 + ⋯ + an . Prove that n+j

j−1

i n+j + xn (1 − x) ∑ an,n+i xi ∑ ai x = an,n+j x i=n

i=0

for all n and j. This is called Abel’s Formula. (b) Use the convergence of ∑ ai and the previous part to show that the power series ∑ ai xi satisfies the Cauchy criterion (Corollary 7.1) on [0, 1]. (10) Use the previous exercise and a change of variable to prove the general form of Abel’s Theorem: Suppose f (x) = ∑ ai xi has radius of convergence 0 < R < ∞ and that it converges at x = R. Then it converges uniformly on [0, R]. Moreover, if it converges at x = −R, then it converges uniformly on [−R, 0].

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(11) Use the previous exercise to prove the statement: Suppose f (x) = ∑ ai xi has radius of convergence 0 < R < ∞ and that it converges at x = R. Then f is continuous at x = R. If it converges at x = −R, then f is continuous at x = −R. (12) Use the previous exercise and the Taylor series at 0 of ln(1 + x) to prove the equality ∞

ln 2 = ∑ (−1)n+1 n=1

(13)

(14)

(15) (16)

1 1 1 1 = 1 − + − + ⋯. n 2 3 4

In other words, the alternating harmonic series converges to ln 2. Suppose {fn } is a decreasing sequence of continuous functions on [a, b] (i.e., fn (x) ≥ fn+1 (x) for all n and x) that converges pointwise on [a, b] to the 0 function. Prove that fn converges to 0 uniformly. Suppose {fn } is an increasing sequence of continuous functions on [a, b] (i.e., fn (x) ≤ fn+1 (x) for all n and x) that converges pointwise to a continuous function f on [a, b]. Prove that fn converges uniformly to f . This result is known as Dini’s Theorem. Give an example to show that the conclusion of the previous exercise may fail if the interval is of the form [a, b). Give an example to show that the increasing assumption cannot be omitted in Dini’s Theorem.

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Section 1.1.4 (1) No. (10) 2n . Section 1.2.5 )n!. (1) No. (6) mn and (m n Section 1.3.7 (20) Pick N with x + N > 0, and apply the x > 0 case to x + N < y + N . Section 1.5.3 (6) Prove this by induction. (14) Try to imitate the proof of Theorem 1.9. (17) Prove this by contradiction. Section 1.6 (7) First prove that S has a countable subset A = {a0 , a1 , . . .} (Exercise (6) on page 6). Define T = S ∖ {a0 }. Let f ∶ S → T be the function defined as ⎧ ⎪ if x ∈/ A, ⎪x f (x) = ⎨ ⎪ ⎪ ⎩an+1 if x = an ∈ A. Prove that f is a bijection. (25) A is bounded above by b1 , and B is bounded below by a1 . To prove α ≤ β, first prove that ai < bj for all i and j. Section 2.1.5 (13) Use any  > 0 with  < L. Section 2.2.4 and that bn > 0 for n > 1. To prove (13) For the first part, note that (n2 ) = n(n−1) 2 bn → 0, first establish 2 > b2n . n−1 (15) For the first part, first prove (1 + r)n > (n2 )r2 . Section 2.3.4 (8) First construct a subsequence that converges to 0. (18) To show that {bn } is decreasing, prove that if S ⊆ T are non-empty and bounded above, then sup(S) ≤ sup(T ). 183

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Section 2.4.5 (11) Use induction to construct a subsequence with ∣ank − lk ∣ < k1 . (16) Since lim sup an = lim sn < M , there exists N such that n ≥ N implies sn < M . Section 2.5 (17) Use an enumeration of the rational numbers between 0 and 1. Section 3.1.5 (13) Use the Monotone Convergence Theorem. (15) The sequence of partial sums for ∑ bn is a subsequence of the sequence of partial sums for ∑ an . Section 3.2.3 (4) Use Cauchy’s criterion and the Triangle Inequality. Section 3.3.1 (2) and (3) Both statements are false. Section 3.4.4 (6) For the first part, use the equality an an aN +1 aN +2 ⋯ = . aN aN +1 an−1 aN 1

For the second part, show that lim sup ann ≤ L + . Section 3.5.1 (2) Prove the first part by contradiction. If there are only finitely many qn , then after excluding finitely many ak we have ∑ an = ∑ pn , which is convergent, and hence absolutely convergent. For the second part, note that {pn } is a subsequence of {an }. The last part is proved by contradiction again. Section 3.6 (1) Use Cauchy Condensation Test. (11) For the last part, use the previous part and the Comparison Test. (12) For the second part, use Kummer’s Test. Section 4.1.1 (5) Prove the “only if” part by proving its contrapositive. Section 4.2.5 (10) If limx→c f = L > 0, use  = L2 . Section 4.3.3 (3) Use the corresponding statements about limits of sequences. Section 4.4.1 (6) Consider the function g(x) = f (x) − x, and use the Intermediate Value Theorem. Section 4.5.1 (7) For the first part, use the previous exercise. For the second part, consider the sequence x1 , y1 , x2 , y2 , . . .. Section 4.6.4 (3) For the first part, if f is not strictly increasing, then there are x < y in

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[a, b] such that f (x) ≥ f (y). So either x =/ a or y =/ b. In the former case, we have a < x < y ≤ b. Now apply the Intermediate Value Theorem on [a, x] for the value f (y) to obtain a contradiction. Section 4.7.4 (12) Use induction on the number of points added to P . Section 4.8 (10) Use the Intermediate Value Theorem on the function g(x) = f (x + 1) − f (x) on [0, 1]. (18) For the first part, pick a sequence of irrational numbers that converges to x. For the second part, first show that every -neighborhood of x contains only finitely many rational numbers pq such that q ≤ 1 . Pick δ such that the δ-neighborhood of x contains no such rational numbers. Section 5.1.6 (2) The sequential definition of f ′ (c) is this: For every sequence {xn } in I ∖ {c} with lim xn = c, the sequence whose nth term is f (xn ) − f (c) xn − c converges to f ′ (c). The -δ definition of f ′ (c) is this: For every  > 0, there exists δ > 0 such that 0 < ∣x − c∣ < δ with x in I implies ∣

f (x) − f (c) − f ′ (c)∣ < . x−c

(8) Use induction and the product rule. (13) No. Section 5.2.4 (2) Consider the function (1 + x)n on [0, a]. (12) Apply the Mean Value Theorem on [a, c] and [c, b]. (14) and (15) First prove the case with y = 1. Then apply this case with x replaced by xy . (16) Use the continuity of f ′ to show the existence of J on which f ′ =/ 0. Then use Rolle’s Theorem. Section 5.3.5 (13) Use the previous exercise, and imitate the proof of L’Hospital’s Rule, using the Mean Value Theorem n + 1 times. Section 5.4 (7) For “only if” imitate the proof of chain rule. Define ⎧ f (x)−f (c) ⎪ ⎪ g(x) = ⎨ ′ x−c ⎪ ⎪ ⎩f (c)

if x =/ c, if x = c.

(9) Apply Rolle’s Theorem to g(x) = e−rx f (x). (11) Set g(x) = x.

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(13) Use Theorem 4.10 and the Intermediate Value Property for derivative. (15) Use the Mean Value Theorem three times, twice for f and once for f ′ . Section 6.1.6 (6) For the first inequality, use f (x) + g(y) ≤ ui (f ) + ui (g) for all x, y in [xi−1 , xi ].  . (12) Choose P with norm < f (b)−f (a) (14) Prove this by induction on n. (15) Use Theorem 6.2, the previous exercise, and an 2 -argument. Section 6.2.3 (4) Use an 2 -argument. (5) The “only if” part is an easy 2 -argument. The other direction is a slightly harder 2 -argument using Theorem 2.10. (7) Use the approximation b

R∫

f ≈ R(f ; P )

a

for P having small norm. Section 6.3.4 (1) To prove that rf is integrable for r > 0, first prove U (rf ; P ) = rU (f ; P )

and L(rf ; P ) = rL(f ; P ).

To prove that rf is integrable for r < 0, first prove U (rf ; P ) = rL(f ; P )

and L(rf ; P ) = rU (f ; P ).

b ∫a f

≥ 0 if f (x) ≥ 0 for all x, first prove U (f ; P ) ≥ 0. To prove (2) To prove that ∣f ∣ is integrable, first prove ui (∣f ∣) − li (∣f ∣) ≤ ui (f ) − li (f ). To prove the inequality, use −∣f ∣ ≤ f ≤ ∣f ∣ and Theorem 6.6. (5) First prove the following statement. If f = 0 except for finitely many points b on [a, b], then f is integrable with ∫a f = 0. (6) Use the Mean Value Theorem for integrals. (8) Prove this by contradiction. If f (x) > 0 for some point x, then there is a small interval containing x on which f ≥ r for some positive real number r. Construct a partition P that includes the end points of this interval, and observe b that L(f ; P ) > 0. Use this to conclude that ∫a f > 0. (9) Note that g is uniformly continuous on [c, d]. We may assume that g is not the 0 function. Thus, given  > 0, there exists 0 < δ < 0 such that  ∣x − y∣ < δ implies ∣g(x) − g(y)∣ < 0 = , 2(b − a) + 4M where M = sup{∣g∣ on [c, d]} > 0. Use the integrability of f to obtain a partition P such that U (f ; P ) − L(f ; P ) < δ 2 .

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187

Then prove that U (g ○ f ; P ) − L(g ○ f ; P ) < . (10) Use the previous exercise with g(x) = x2 . (11) First prove f g = 41 [(f + g)2 − (f − g)2 ]. (12) Start with the inequality mg(x) ≤ f (x)g(x) ≤ M g(x), where m = inf{f (x) ∶ x ∈ [a, b]} and M = sup{f (x) ∶ x ∈ [a, b]}. Integrate and use the Intermediate Value Theorem for f . Section 6.4.7 (1) Use Theorem 6.12 and take derivative on both sides to obtain f (x) = −f (x). x (2) First write g in terms of F (x) = ∫0 f . Section 6.5 (1) First compute derivatives on both sides. (3) A function of bounded variation is the difference of two increasing functions (Theorem 4.16), which are integrable by Exercise (12) on page 141. (6) For the first part, use Exercise (7) on page 98. For the second part, use Exercise (5) on page 149 and Theorem 6.7. (7) Use Exercise (13) on page 13, Exercise (5) on page 5, and Theorem 6.7. (8) For the “if” part, take a large enough closed sub-interval of (a, b) and use b b−δ an 3 -argument. ∫a f = limδ→0+ ∫a+δ f (9) Use Theorem 6.2, Theorem 6.4, and an 3 -argument. (10) Use the step function in (4.5). (12) Take f as the Thomae function on [0, 1] (Exercise (14) on page 149). Define g as g(0) = 0 and g(x) = 1 for x > 0. Section 7.1.6 (12) Use the approximations and inequality f (x) ≈ fn (x) ≤ fn (y) ≈ f (y) for x < y. (13) Approximate f (a) with fn (a) and then with fn (an ). Section 7.2.5 (3) Use an argument similar to the proof of the product rule. (4) Enumerate the rational numbers in [0, 1] as q1 , q2 , . . .. Define ⎧ ⎪ ⎪1 if x = q1 , ..., qn , fn (x) = ⎨ ⎪ ⎪ ⎩0 otherwise. Show that fn is integrable, and that f is the characteristic function on Q. (12) Try fn = n1 χQ . Section 7.3.6 (3) Justify and use the equalities

i=1

n

b

∞

∑ (∫

a

n→∞

i=1

b n

b

fi ) = lim ∑ ∫

a

fi = lim ∫ n→∞

b

∑ fi = ∫

a i=1

n

b

lim ∑ fi = ∫

a n→∞ i=1

a

f.

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(8) Use Corollary 7.1 with n = m + 1. Section 7.4.7 (3) The Weierstrass M -Test would not apply when Mn = 1n . (8) Prove by induction that ⎧ ⎪ if x = 0, ⎪0 f (n) (x) = ⎨ −1/x2 −3n ⎪ x Pn (x) if x =/ 0, ⎪ ⎩e where Pn is some non-zero polynomial of degree < 3n. Section 7.5 (5) Use an 4 -argument and the approximations b

∫

a

b

f ≈ U (f ; P ) ≈ U (fn ; P ) ≈ ∫

a

xn

fn ≈ ∫

a

fn .

(6) Try fn = 21n xχQ . (8) For the first part, use the approximations f (x) ≈ fn (x) ≈ fn (y) ≈ f (y). For the second part, use the approximations fn (x) ≈ f (x) ≈ f (y) ≈ fn (y). (9) For the first part, expand the right-hand side in terms of ak . For the second part, first use the Cauchy criterion on the convergent series ∑ ai to conclude ∣am,n ∣ <  for large m and any n ≥ m. (11) Reduce to the situation of the previous exercise by considering g(x) = f (Rx).

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List of Notations

Notation x∈S y ∈/ S S⊆T S ∪ T, S ∩ T S ∖ T, S × T Dom(f ) Ran(f ) f (A) f −1 (B) f −1 g○f sup(S) inf(S) n! (nk) {an } lim an = L, a → L lim an = ±∞, an → ±∞ {ank } lim sup lim inf ∞ ∑n=1 an , ∑ an limx→c f limx→c− f limx→c+ f χS {a = x0 < ⋯ < xn = b} V (f ; P ) V (f ; [c, d]) df f ′ , dx

Page 1 1 2 2 3 4 4 4 5 6 6 11 11 18 18 30 31 33 45 53 53 62 85 88 89 93 105 105 105 115, 116

Description x is an element in S y is not an element in S S is a subset of T union, intersection difference, product domain of f range of f image of A under f inverse image of B under f inverse function composition supremum of S infimum of S n factorial binomial coefficient sequence sequence converges to L sequence diverges to ±∞ subsequence limit superior limit inferior series limit of f left-hand limit right-hand limit characteristic function of S partition of [a, b] variation of f with respect to P variation of f on [c, d] derivative 189

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f (n) Tn (x; c) Rn (x; c) ∣∣P ∣∣ ∆xi li (f ) ui (f ) L(f ; P ) U (f ; P ) L(f ) U (f ) b b ∫a f , ∫a f (x)dx R(f ; P ) b R ∫a f

115 126 126 136 136 136 136 136 136 136 136 138 142 143

nth derivative degree n Taylor polynomial nth error term norm of P xi − xi−1 in P infimum of f on [xi−1 , xi ] supremum of f on [xi−1 , xi ] lower sum of f with respect to P upper sum of f with respect to P lower sum of f upper sum of f integral of f on [a, b] Riemann sum of f with respect to P Riemann integral of f on [a, b]

f ∣a limn→∞ fn fn → f pointwise fn → f uniformly ≈ ∑ fn ∑ ai (x − c)i T (f ; c)(x)

151 158 158 159 163 169 175 179

f (b) − f (a) pointwise limit of {fn } {fn } converges pointwise to f {fn } converges uniformly to f approximation series of functions power series at c Taylor series of f at c

b

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Bibliography

Gelbaum, B. R. and Olmsted, J. M. H. (1964). Counterexamples in Analysis (Holden-Day, San Francisco). Halmos, P. R. (1974). Naive Set Theory (Springer, New York). Hobson, E. W. (1907). The Theory of Functions of a Real Variable and the Theory of Fourier’s Series (Cambridge University Press, Cambridge). Rudin, W. (1976). Principles of Mathematical Analysis (McGraw-Hill, New York). Sprecher, D. A. (1987). Elements of Real Analysis (Dover, New York).

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Index

Abel’s Formula, 82, 181 Abel’s Test, 82 Abel’s Theorem, 80, 181 algebra of functions, 109 Alternating Series Test, 70 Archimedean Property, 12

right, 113 uniformly, 96 convergence function, 85 sequence, 31 series, 62

Bernoulli’s Inequality, 18 binomial coefficient, 18 Binomial Theorem, 19 Bolzano-Weierstrass Theorem, 46 bounded, 10 above, 10 below, 10 bounded variation, 105

derivative, 115 left-hand, 132 right-hand, 132 differentiable, 115 continuously, 116 infinitely, 116 Dini’s Theorem, 182 Dirichlet’s Test, 82 discontinuous, 91 divergence sequence, 31, 33 series, 62

Cantor’s diagonal argument, 23 Carath´eodory’s Theorem, 132 Cauchy Condensation Test, 64 Cauchy criterion integrability, 141 Riemann integrability, 145 sequence, 48 series, 63 series of functions, 170 Cauchy product, 82 Chain Rule, 118 characteristic function, 93 Comparison Test, 67 Completeness Axiom, 11 continuous, 91 -δ characterization, 92 extension, 114 left, 112 nowhere differentiable, 172 piecewise, 155

 -argument, 2

35 equicontinuous, 181 error term, 127 extended real number, 51 Extreme Value Theorem, 94 factorial, 18 field, 8 ordered, 9 function, 4 bijection, 5 bounded, 93 bounded variation, 105 characteristic, 93 composition, 6 decreasing, 99 193

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domain, 4 even, 132 image, 4 increasing, 99 injection, 5 inverse, 6, 101 inverse image, 5 limit, 85 Lipschitz, 112 monotone, 99 odd, 132 periodic, 111, 172 periodic extension, 112, 174 piecewise continuous, 155 piecewise monotone, 155 range, 4 step, 104 strictly decreasing, 99 strictly increasing, 99 strictly monotone, 99 surjection, 5 target, 4 Thomae, 112, 149 Fundamental Theorem of Calculus, 150, 153 induction, 17, 20 infimum, 11 integrable, 138 -criterion, 139 integral, 138 improper, 156 linearity, 146 Integral Test, 156 integration by parts, 151 interchange of limits, 163, 171 Interior Extremum Theorem, 122 Intermediate Value Theorem, 94 for derivative, 129 interval, 14 interval of convergence, 176 Inverse Function Theorem, 119 irrational number, 2, 13 isolated point, 111 Kummer’s Test, 81 L’Hospital’s Rule, 128 limit -δ characterization, 87

at ∞, 110 function, 85 inferior, 53 left-hand, 88 right-hand, 89 sequence, 31 series, 62 subsequential, 52 superior, 53, 175 uniqueness, 34 Limit Comparison Test, 69 limit point, 83 lower bound, 10 greatest, 11 lower sum, 136 with respect to P , 136 Mean Value Theorem, 123 Cauchy, 132 for integral, 148 Monotone Convergence Theorem, 39, 47 Nested Interval Property, 27, 44 norm, 136 partial sum, 62, 169 partition, 105 point, 105 tagged, 142 Pascal’s Triangle, 18 peak, 46 pointwise convergence, 157 pointwise limit, 158 power series, 175 term-wise differentiation, 178 term-wise integration, 177 uniform convergence, 177 Product Rule, 117 Generalized, 121 Quotient Rule, 117 Raabe’s Test, 81 radius of convergence, 175 Ratio Test, 74 rational number, 2, 13 countable, 22 refinement, 136 Riemann integrable, 143 Riemann integral, 143

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Index

Riemann Rearrangement Theorem, 79 Riemann sum, 142 Rolle’s Theorem, 123 Root Test, 73 sequence, 30 bounded, 37 Cauchy, 47 contractive, 50 decreasing, 38 Fibonacci, 30 increasing, 38 monotone, 38 strictly decreasing, 38 strictly increasing, 38 strictly monotone, 38 subsequence, 45 sequence of functions, 157 series, 62 p-series, 65 absolutely convergent, 72 alternating, 70 alternating harmonic, 70 conditionally convergent, 72 geometric, 30, 64 harmonic, 64 rearrangement, 77 series of functions, 169 set, 1 closed, 26, 113 countable, 22 difference, 3 disjoint, 3 disjoint union, 3 empty, 1 finite, 21 infinite, 21 intersection, 3

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195

non-empty, 1 open, 26, 113 power, 4, 26 product, 3 proper subset, 2 subset, 2 symmetric difference, 25 uncountable, 22 union, 2 Squeeze Theorem, 42 subsequence, 45 limit, 52 substitution, 154 supremum, 11 Taylor polynomial, 127 Taylor series, 179 Taylor’s Theorem, 127 integral form, 152 telescoping sum, 49 Tietze Extension Theorem, 114 Triangle Inequality, 9 unbounded, 10 uniform convergence, 159 uniform limit, 159 uniformly bounded, 169 uniformly Cauchy, 160 upper bound, 10 least, 11 upper sum, 136 with respect to P , 136 variation, 105 Weierstrass M -Test, 170 Well-Ordering Property, 17